Question

Hello. How can I solve cos(γ+135°)+cos(γ-135°)?

Answer 1

try using the sum and difference identities for cosine
also i don't think the instructions say solve

Answer 2

\[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) \\ \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\]

Answer 3

Everything I was able to do with it was this:
cos(γ+135°)+cos(γ-135°)
= cosγ*cos135°-sinγ*sin135°+cosγ*cos135°+sinγ*sin135°
=cosγ*cos135°*sin²135°

and the book says I'm supposed to get this: -√2cosγ

Answer 4

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

cos(y+135)+cos(y-135)
= (cosy*cos135- siny*sin135)+(cosy*cos135+siny*sin135)

Put value of cos135 and sin135
or you can further expand it by making 135 = 90+45

Answer 5

I have given the solution, do check it

Answer 6

\[\cos( \gamma + 135^o) + \cos( \gamma - 135^o)\]
\[(\cos \gamma \cos 135^o - \sin \gamma \sin 135^o) + (\cos \gamma \cos 135^o +\sin \gamma \sin135^o)\]
removing the parenthesis:
\[\cos \gamma \cos 135^o - \sin \gamma \sin 135^o + \cos \gamma \cos 135^o+\sin \gamma \sin135^o\]
\[\sin \gamma \sin 135^o - \sin \gamma \sin 135^o \rightarrow 0\]
\[\cos \gamma \cos 135^o + \cos \gamma \cos 135^o\]
\[2\cos \gamma \cos 135^o\]
\[\cos 135^o \rightarrow \frac{ -\sqrt{2} }{ 2 }\]
\[2 \frac{ -\sqrt{2} }{ 2 } \cos \gamma \rightarrow -\sqrt{2}\cos \gamma\]

Answer 7

thank you (: