Trigonometry Problems - sin, cos, tan, cot: Very Difficult Problems with Solutions

Solution:
[tex]x+y+z=\pi[/tex], so [tex]{\frac{x}{2}}+{\frac{y}{2}}={\frac{\pi}{2}}-{\frac{z}{2}}; {\frac{z}{2}}={\frac{\pi}{2}}-({\frac{x}{2}}+{\frac{y}{2}})[/tex]. Because of [tex]cot\alpha={\frac{1}{tan\alpha}}[/tex], we get [tex]cot\alpha={\frac{cos\alpha}{sin\alpha}}[/tex]. Then [tex]cot{\frac{x}{2}}+cot{\frac{y}{2}}+cot{\frac{z}{2}}={\frac{cos{\frac{x}{2}}}{sin{\frac{x}{2}}}}+{\frac{cos{\frac{y}{2}}}{sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}={\frac{cos{\frac{x}{2}}sin{\frac{y}{2}}+cos{\frac{y}{2}}sin{\frac{x}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}={\frac{sin({\frac{x}{2}}+{\frac{y}{2}})}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}={\frac{cos{\frac{z}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}=cos{\frac{z}{2}}\cdot{\frac{sin{\frac{z}{2}}+sin{\frac{x}{2}}sin{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}sin{\frac{z}{2}}}}={\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}\cdot{\frac{cos({\frac{x}{2}}+{\frac{y}{2}})+sin{\frac{x}{2}}sin{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}={\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}\cdot{\frac{cos{\frac{x}{2}}cos{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}=cot{\frac{x}{2}}cot{\frac{y}{2}}cot{\frac{z}{2}}[/tex].

Solution:
5cosA +12sinA + 12 = 13(5/13 cosA +12/13sinA) + 12
Now, for any values of B we can get sinB = 5/13 and we can replace cosB = 12/13.
We see that our assumption is right because we satisfy the condition sin^2B + cos^2B = 1 so we get 13(sinBcosA +cosBsinA) + 12 =13(sin(A+B))+12.
Therefore we know that minimum value of sinx=-1 and greatest is 1. Тhe greatest value is when sin(A + B) = 1 then value of the expression becomes 13.1 + 12 = 25

Solution:
5cosA +12sinA + 12 = 13(5/13 cosA +12/13sinA) + 12
Now, for any values of B we can get sinB =5/13 and we can replace cosB = 12/13.
We see that our assumption is right because we satisfy the condition sin^2B + cos^2B=1 so we get 13(sinBcosA +cosBsinA) + 12 =13(sin(A+B))+12. Therefore we know that minimum value of sinx=-1 and greatest is 1 so the least value of the expression becomes when sin(A+B) =-1 then the value of the whole term becomes 13.(-1) + 12 = -13 +12 = -1

Solution:
[tex]\sin A = 1 - \sin^2 A[/tex]
[tex]\sin A = \cos^2 A[/tex]
[tex]\sin^2 A = \cos^4 A[/tex]
[tex]1 - \cos ^2 A = \cos^4 A[/tex]
[tex]1 = \cos^4 A + \cos^2 A[/tex]
[tex]1^3 = (\cos^4 A + \cos^2 A)^3[/tex] by formula [tex](a+b)^3[/tex]
[tex]1 = \cos^{12} A + 3 \cos^{10} A +3 \cos^8 A + \cos^6 A [/tex]
Transverse 1 onto the other side we get the condition as given as in the question. Comparing the variables we get [tex]a = 1[/tex], [tex]b = 3[/tex], [tex]c = 3[/tex], [tex]d = 1[/tex] hence the value of [tex]b +\frac{c}{a}+b = \frac{3 + 3}{1 + 1} = \frac{6}{2} = 3[/tex]