Calculate sin (alpha - beta) and cos (alpha + beta)

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

Calculate sin (alpha - beta) and cos (alpha + beta)

Postby Guest » Thu Oct 18, 2018 6:59 pm

Calculate sin (alpha - beta) and cos (alpha + beta)

IF:

cos alpha = - 5/13
sin beta = 12/13

pi/2 < alpha < pi, pi/2 < beta < pi
Guest
 

Re: Calculate sin (alpha - beta) and cos (alpha + beta)

Postby Guest » Wed Aug 28, 2019 7:52 am

I thought it was well known that
[tex]cos(\alpha+ \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)[/tex] and
[tex]sin(\alpha+ \beta)= sin(\alpha)cos(beta)+ sin(\beta)cos(\alpha)[/tex]
(https://brownmath.com/twt/sumdiff.htm)

Now replace [tex]\beta[/tex] with [tex]-\beta[/tex] and use the fact that cos(-x)= cos(x) and sin(-x)= -sin(x).
Guest
 

Re: Calculate sin (alpha - beta) and cos (alpha + beta)

Postby Guest » Fri Jan 10, 2020 1:51 am

[tex]Cos(\alpha)=\frac{-5}{13}\Rightarrow Sin^{2}(\alpha)+ Cos^{2}(\alpha)=1 \Rightarrow Sin(\alpha)=\frac{12}{13}[/tex]
[tex]Sin(\beta)=\frac{12}{13}\Rightarrow Sin^{2}(\beta)+ Cos^{2}(\beta)=1 \Rightarrow Cos(\beta)=\frac{5}{13}[/tex]
[tex]Sin(\alpha-\beta)=Sin(\alpha)Cos(\beta)-Cos(\alpha)Sin(\beta) \Rightarrow Sin(\alpha-\beta)=2(\frac{12}{13})(\frac{5}{13})=\frac{120}{169}[/tex]
[tex]Cos(\alpha+\beta)=Cos(\alpha)Cos(\beta)-Sin(\alpha)Sin(\beta) \Rightarrow Cos(\alpha+\beta)=-\frac{25}{169}-\frac{144}{169}=-1[/tex]
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