Trigonometric Identities: Problems with Solutions

Solution:
Answer: $\frac{2}{\sin x}=\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}$

$\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{\sin ^{2}x+\left( 1+\cos x\right) ^{2}}{\sin x\left( 1+\cos x\right) }=\frac{\sin ^{2}x+1+2\cos x+\cos ^{2}x}{\sin x\left( 1+\cos x\right) }=\frac{1+2\cos x+\left( \sin ^{2}x+\cos ^{2}x\right) }{\sin x\left( 1+\cos x\right) }$

So $=\frac{2+2\cos x}{\sin x\left( 1+\cos x\right) }=\frac{2\left( 1+\cos x\right) }{\sin x\left( 1+\cos x\right) }=\frac{2}{\sin x}$

Then $\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x}$

Solution:
Answer: $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$

$\frac{\cos x}{1+\sin x}=\frac{\cos ^{2}x}{\cos x\left( 1+\sin x\right) }=\frac{1-\sin ^{2}x}{\cos x\left( 1+\sin x\right) }=\frac{\left( 1-\sin x\right) \left( 1+\sin x\right) }{\cos x\left( 1+\sin x\right) }=\frac{1-\sin x}{\cos x}$

We proved that $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$

Solution:
Answer: $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\tan x-1}{\tan x+1}$

$\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\frac{1}{\cos x}-\frac{1}{\sin x}}{\frac{1}{\cos x}+\frac{1}{\sin x}}=\frac{\frac{\sin x-\cos x}{\sin x\cos x}}{\frac{\sin x+\cos x}{\sin x\cos x}}=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{\sin x+\cos x}{\cos x}}=\frac{\frac{\sin x}{\cos x}-1}{\frac{\sin x}{\cos x}+1}=\frac{\tan x-1}{\tan x+1}$

We proved that $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\tan x-1}{\tan x+1}$

Solution:
Answer: $\frac{\tan x-\sin x}{\sin^{3}x}=\frac{1}{\cos x+\cos^{2}x}$

$\frac{\tan x-\sin x}{\sin^{3}x}=\frac{\frac{\sin x}{\cos x}-\sin x}{\sin ^{3}x}=\frac{\sin x-\sin x\cos x}{\cos x\sin ^{3}x}=\frac{\sin x\left( 1-\cos x\right) }{\cos x\sin^{3}x}$
$= \frac{1-\cos x}{\cos x\sin^{2}x}=\frac{1-\cos x}{\cos x\left( 1-\cos ^{2}x\right) }=\frac{1}{\cos x\left( 1+\cos x\right) }=\frac{1}{\cos x+\cos ^{2}x}$

We proved that $\frac{\tan x-\sin x}{\sin ^{3}x}=\frac{1}{\cos x+\cos^{2}x}$

Solution:
Answer: $\frac{\cos ^{3}x-\sin ^{3}x}{\cos x-\sin x}=1+\sin x\cos x$

$\frac{\cos ^{3}x-\sin^{3}x}{\cos x-\sin x} =\frac{\left( \cos x-\sin x\right) \left( \cos^{2}x+\cos x\sin x+\sin ^{2}x\right) }{\cos x-\sin x}=\left( \cos ^{2}x+\sin ^{2}x\right) +\cos x\sin x=$
$=1+\cos x\sin x$

We proved that $\frac{\cos^{3}x-\sin^{3}x}{\cos x-\sin x}=1+\sin x\cos x$

Solution:
Answer: $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{\sin \theta +1}{\cos \theta }$

$\frac{\sin \theta +1\ }{\cos \theta }=\frac{\left( \sin \theta +1\right) \left( \sin \theta +\cos \theta -1\right) }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }=\frac{\sin ^{2}\theta +\sin \theta \cos \theta +\cos \theta -1}{\cos \theta \left( \sin \theta +\cos \theta -1\right) }= \frac{\left( \sin ^{2}\theta -1\right) +\sin \theta \cos \theta +\cos \theta }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }$

$=\frac{-\cos^{2}\theta +\sin \theta \cos \theta +\cos \theta }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }=\frac{\cos \theta \left( \sin \theta -\cos \theta +1\right) }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }=\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$
We proved that $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{\sin \theta +1\ }{\cos \theta }$

Solution:
Answer: $\tan \left( \alpha +\beta \right) =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$\tan \left( \alpha +\beta \right) =\frac{\sin \left( \alpha +\beta \right) }{\cos \left( \alpha +\beta \right) }=\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }$

Divide the numerator and denominator by $\cos \alpha \cos \beta $

So $\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }=\frac{\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}=\frac{\frac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta }+\frac{\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

We proved that $\tan \left( \alpha +\beta \right) =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

Solution:
Answer: $\sin 15^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) \quad \cos 15^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) \quad \tan 15^{\circ }=2-\sqrt{3}$

$\sin 15^{\circ }=\sin \left( 45^{\circ }-30^{\circ }\right) =\sin 45^{\circ }\cos 30^{\circ }-\cos 45^{\circ }\sin 30^{\circ }=\tfrac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) $

$\cos 15^{\circ }=\cos \left( 45^{\circ }-30^{\circ }\right) =\cos 45^{\circ }\cos 30^{\circ }+\sin 45^{\circ }\sin 30^{\circ }=\tfrac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) $

$\tan 15^{\circ }=\tan \left( 45^{\circ }-30^{\circ }\right) =\frac{\tan 45^{\circ }-\tan 30^{\circ }}{1+\tan 45^{\circ }\tan 30^{\circ }}=\frac{1-\frac{1}{\sqrt{3}}}{1+1\left( \frac{1}{\sqrt{3}}\right) }=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

Solution:
Answer: $\sin 75^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) \quad \cos 75^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) \quad \tan 75^{\circ }=\frac{\left( \sqrt{3}+1\right) ^{2}}{2}$

$\sin 75^{\circ }=\sin \left( 90^{\circ }-15^{\circ }\right) =\sin 90^{\circ }\cos 15^{\circ }-\cos 90^{\circ }\sin 15^{\circ }=$

$1\cdot \frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) -0\cdot \frac{\sqrt{2}}{4}% \left( \sqrt{3}-1\right) =\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) $

$\cos 75^{\circ }=\cos \left( 90^{\circ }-15^{\circ }\right) =\cos 90^{\circ }\cos 15^{\circ }+\sin 90^{\circ }\sin 15^{\circ }=$

$0\cdot \frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) +1\cdot \frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) =\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) $

$\tan 75^{\circ }=\frac{\sin 75^{\circ }}{\cos 75^{\circ }}=\frac{\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) }{\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) }=\frac{\left( \sqrt{3}+1\right)^{2}}{\left( \sqrt{3}-1\right) \left( \sqrt{3}+1\right) }=\frac{\left( \sqrt{3}+1\right)^{2}}{2}$

Solution:
$\sin \left( \alpha +\beta \right) +\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta +\sin \beta \cos \alpha +\sin \alpha \cos \beta -\sin \beta \cos \alpha$

Then $\sin \left( \alpha +\beta \right) +\sin \left( \alpha -\beta \right) =2\sin \alpha \cos \beta$

Solution:
$\sin \left( \alpha +\beta \right) -\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta +\sin \beta \cos \alpha -\left( \sin \alpha \cos \beta-\sin \beta \cos \alpha \right) $
$=2\cos \alpha \sin \beta $

Solution:
$\cos \left( \alpha +\beta \right) +\cos \left( \alpha -\beta \right) =\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right) +\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right) =2\cos \alpha \cos \beta$

Solution:
$\cos \left( \alpha +\beta \right) -\cos \left( \alpha -\beta \right) =\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right) -\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right) =-2\sin \alpha \sin \beta $

Solution:
Using the identity $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$, we get
$\frac{\tan \left( \alpha +\beta \right) -\tan \alpha }{1+\tan \left( \alpha +\beta \right) \tan \alpha }=\tan \left[ \left( \alpha +\beta \right) -\alpha \right] =\tan \beta $

We proved that $\frac{\tan \left( \alpha +\beta \right) -\tan \alpha }{1+\tan \left( \alpha +\beta \right) \tan \alpha }=\tan \beta $

Solution:
Since $\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta -\cos \alpha \sin \beta $ and $\cos \left( \alpha -\beta \right) =\cos \alpha \cos \beta +\sin \alpha \sin \beta $
Then
$\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right) ^{2}+\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right) ^{2}=\sin ^{2}\left( \alpha -\beta \right) +\cos ^{2}\left( \alpha -\beta \right) =1$

Solution:
$\cot \left( \alpha +\beta \right) =\frac{1}{\tan \left( \alpha +\beta \right) }=\frac{1-\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }=\frac{1-\frac{1}{\cot \alpha \cot \beta }}{\frac{1}{\cot \alpha }+\frac{1}{\cot \beta }}=\frac{\frac{\cot \alpha \cot \beta -1}{\cot \alpha \cot \beta }}{\frac{\cot \beta +\cot \alpha }{\cot \alpha \cot \beta }}=\frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }$

We proved that $\cot \left( \alpha +\beta \right ) =\frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }$

Solution:
Answer: $\cot \left( \alpha -\beta \right) =\frac{\cot \alpha \cot \beta +1}{\cot \beta -\cot \alpha }$

Since $\cot \left( -\beta \right) =-\cot \left( \beta \right) $
and, if we use that $\cot \left( \alpha +\beta \right) =\frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }$

Then $\cot \left( \alpha -\beta \right) =\cot \left[ \alpha +\left( -\beta \right) \right] =\frac{\cot \alpha \cot \left( -\beta \right) -1}{\cot \left( -\beta \right) +\cot \alpha }=\frac{-\cot \alpha \cot \beta -1}{-\cot \beta +\cot \alpha }=\frac{\cot \alpha \cot \beta +1}{\cot \beta -\cot \alpha }$

We proved that $\cot \left( \alpha +\beta \right) =\frac{\cot \alpha \cot \beta +1}{\cot \beta -\cot \alpha }$

Solution:
We know that $\cos 2\alpha =\cos^{2}\alpha -\sin ^{2}\alpha =\left( \cos ^{2}\alpha +\sin ^{2}\alpha \right) -2\sin ^{2}\alpha =1-2\sin ^{2}\alpha $,

Let $\alpha =\frac{1}{2}\theta $.
Then, $\cos \theta =1-2\sin ^{2}\frac{1}{2}\theta $

$\sin ^{2}\frac{1}{2}\theta =\frac{1-\cos \theta }{2}$
So

$\sin \frac{1}{2}\theta =\pm \sqrt{\frac{1-\cos \theta }{2}}$

Solution:
$\cos 2\alpha =\cos^{2}\alpha -\sin^{2}\alpha =\left( \cos^{2}\alpha +\cos^{2}\alpha \right) -\cos^{2}\alpha -\sin ^{2}\alpha =\left( \cos^{2}\alpha +\cos ^{2}\alpha \right) -\left( \cos ^{2}\alpha +\sin^{2}\alpha \right) $
$=2\cos ^{2}\alpha -1$,
Let $\alpha =\frac{1}{2}\theta $.

$\cos \theta =2\cos ^{2}\frac{1}{2}\theta -1$,then $\cos ^{2}\frac{1}{2}\theta =\frac{1+\cos \theta }{2}$

So $\cos \frac{1}{2}\theta =\pm \sqrt{\frac{1+\cos \theta }{2}}$

Solution:
$\tan \frac{1}{2}\theta =\frac{\sin \frac{1}{2}\theta }{\cos \frac{1}{2}\theta }=\frac{\pm \sqrt{\frac{1-\cos \theta }{2}}}{\pm \sqrt{\frac{1+\cos \theta }{2}}}=\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\pm \sqrt{\frac{\left( 1-\cos \theta \right) \left(1+\cos \theta \right) }{\left( 1+\cos \theta \right) \left( 1+\cos \theta \right) }}=\pm \sqrt{\frac{1-\cos ^{2}\theta }{\left( 1+\cos \theta \right) ^{2}}}=\frac{\sin \theta }{1+\cos \theta }$

We proved that $\tan \frac{1}{2}\theta =\frac{\sin \theta }{1+\cos \theta }$

Solution:
Answer: $1-\frac{1}{2}\sin 2x=\frac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}$

When we factor the numerator we get

$\frac{\sin ^{3}x+\cos^{3}x}{\sin x+\cos x}=\frac{\left( \sin x+\cos x\right) \left( \sin^{2}x-\sin x\cos x+\cos ^{2}x\right) }{\sin x+\cos x}=\sin ^{2}x-\sin x\cos x+\cos ^{2}x=$
$=1-\sin x\cos x=1-\frac{1}{2}\left( 2\sin x\cos x\right) =1-\frac{1}{2} \sin 2x$

We proved that $1-\frac{1}{2}\sin 2x=\frac{\sin^{3}x+\cos ^{3}x}{\sin x+\cos x}$

Solution:
Since $\sin \left( \theta +30^{\circ }\right) +\cos \left( \theta +60^{\circ }\right) =\left( \sin \theta \cos 30^{\circ }+\cos \theta \sin 30^{\circ}\right) +\left( \cos \theta \cos 60^{\circ }-\sin \theta \sin 60^{\circ}\right) $

Now we replace $\cos 30^{\circ },\sin 30^{\circ },\cos 60^{\circ },\sin 60^{\circ }$

$\sin \left( \theta +30^{\circ }\right) +\cos \left( \theta +60^{\circ }\right) =\frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta +\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta =\cos \theta $

Solution:
$\frac{1-\tan^{2}\frac{1}{2}x}{1+\tan^{2}\frac{1}{2}x}=\frac{1-\frac{\sin ^{2}\frac{1}{2}x}{\cos^{2}\frac{1}{2}x}}{\sec^{2}\frac{1}{2}x}=\frac{\left( 1-\frac{\sin^{2}\frac{1}{2}x}{\cos ^{2}\frac{1}{2}x}\right) \cos^{2}\frac{1}{2}x}{\sec^{2}\frac{1}{2}x\cos ^{2}\frac{1}{2}x}=\cos ^{2}\frac{1}{2}x-\sin^{2}\frac{1}{2}x=\cos x$