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Trigonometric Ratios of Some Particular Angles — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsTrigonometric Ratios of Some Particular Angles4 Marks
Question
Evaluate the following:
If $A=60^{\circ}$ and $B=30^{\circ}$, verify that:
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
✓

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
Now, $\text{A}+\text{B}=60^\circ+30^\circ=90^\circ$
Also, $\text{A}-\text{B}=60^\circ-30^\circ=30^\circ$
$\cos(\text{A}+\text{B})=\cos90^\circ=0$
$\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
$=\cos60^\circ\cos30^\circ-\sin60^\circ\sin30^\circ$
$=\Big(\frac{1}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times\frac12\Big)=\Big(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\Big)=0$
$\therefore\ \cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$

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