sum and difference identities
- Thread starter ShanikaC
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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.
If you get stuck, please reply showing all the steps and formulas you have tried. Thank you.
Eliz.
Edit: Ne'mind; answer posted below.
Hello, ShanikaC!
\(\displaystyle \;\;\sin(A\,\pm\,B)\:=\:\sin(A)\cdot\cos(B)\,\pm\,\sin(B)\cdot\cos(A)\)
\(\displaystyle \;\;\cos(A\,\pm\,B)\:=\;\cos(A)\cdot\cos(B)\,\mp\,\sin(A)\cdot\sin(B)\)
\(\displaystyle \;\;\tan(A\,\pm\,B)\:=\:\frac{\tan(A)\,\pm\,\tan(B)}{1\,\mp\,\tan(A)\cdot\tan(B)}\)
Since \(\displaystyle 15^o\:=\:45^o\,-\,30^o\), we can the first one like this:
\(\displaystyle \L\sin(45^o\,-\,30^o)\;=\;\sin(45^o)\cdot\cos(30^o)\,-\,\sin(30^o)\cdot\cos(45^o)\)
. . . . . . \(\displaystyle \L= \;\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\,-\,\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) \;= \;\frac{\sqrt{6}\,-\,\sqrt{2}}{4}\)
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:
\(\displaystyle \L\sin(45^o\,-\,30^o)\;=\;\sin(45^o)\cdot\cos(30^o)\,-\,\sin(30^o)\cdot\cos(45^o)\)
. . . . . . \(\displaystyle \L= \;\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\,-\,\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) \;= \;\frac{\sqrt{6}\,-\,\sqrt{2}}{4}\)