sum and difference identities

[ShanikaC]

find exact vaules of sin cos and tanof the given angle: theta=15degree using sum and difference identities

 

[stapel]

Hint: Find a basic "reference" angle that you can relate to "15°" by using sums and/or differences. Then apply the appropriate formula(s).

If you get stuck, please reply showing all the steps and formulas you have tried. Thank you.

Eliz.

Edit: Ne'mind; answer posted below.

 

[soroban]

Hello, ShanikaC!

Find exact vaules of $\displaystyle sin,\;\cos,\;\tan$ of $\displaystyle 15^o$ using sum and difference identities

You are expected to know those sum and difference identities:

$\displaystyle \;\;\sin(A\,\pm\,B)\:=\:\sin(A)\cdot\cos(B)\,\pm\,\sin(B)\cdot\cos(A)$

$\displaystyle \;\;\cos(A\,\pm\,B)\:=\;\cos(A)\cdot\cos(B)\,\mp\,\sin(A)\cdot\sin(B)$

$\displaystyle \;\;\tan(A\,\pm\,B)\:=\:\frac{\tan(A)\,\pm\,\tan(B)}{1\,\mp\,\tan(A)\cdot\tan(B)}$


Since $\displaystyle 15^o\:=\:45^o\,-\,30^o$, we can the first one like this:

\(\displaystyle \L\sin(45^o\,-\,30^o)\;=\;\sin(45^o)\cdot\cos(30^o)\,-\,\sin(30^o)\cdot\cos(45^o)\)

. . . . . . \(\displaystyle \L= \;\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\,-\,\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) \;= \;\frac{\sqrt{6}\,-\,\sqrt{2}}{4}\)

 

[happy]

What if Shanika wanted to try it on her own with the hints that were provided by Stapel?

 

[Denis]

What if Shanika wanted to try it on her own with the hints that were provided by Stapel?

All dear Shanika needs to do then is NOT LOOK :idea: