Question

Find the value of the expression given below,
$\sin 30{}^{\circ }\cos 60{}^{\circ }+\sin 60{}^{\circ }\cos 30{}^{\circ }$
[a] ${\displaystyle \frac{1}{2}}$
[b] ${\displaystyle \frac{\sqrt{3}}{2}}$
[c] 1
[d] ${\displaystyle \frac{1}{4}}$

Answer 1

Hint: Substitute the values of $\sin 30{}^{\circ },\sin 60{}^{\circ },\cos 30{}^{\circ },\cos 60{}^{\circ }$ and simplify and hence find the value of the given expression

Complete step-by-step answer:
To solve the given question, we need to remember the values of $\sin 30{}^{\circ },\sin 60{}^{\circ },\cos 30{}^{\circ }$ and $\cos 60{}^{\circ }$
Consider the following tables of values of sine, cosine, tangent, cotangent, secant and cosecant for angles of measure $0{}^{\circ },30{}^{\circ },45{}^{\circ },60{}^{\circ },90{}^{\circ }$

From the above table, we have
$\sin 30{}^{\circ }={\displaystyle \frac{1}{2}},\sin 60{}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}},\cos 30{}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}},\cos 60{}^{\circ }={\displaystyle \frac{1}{2}}$
Substituting the values of $\sin 30{}^{\circ },\sin 60{}^{\circ },\cos 30{}^{\circ },\cos 60{}^{\circ }$, we get
$S={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}\times {\displaystyle \frac{\sqrt{3}}{2}}$
Simplifying, we get
$S={\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{4}}={\displaystyle \frac{4}{4}}=1$
Hence, we have
$\sin 30{}^{\circ }\cos 60{}^{\circ }+\sin 60{}^{\circ }\cos 30{}^{\circ }$ is equal to 1.
Therefore the correct answer is option (c).

Note: [1] Alternatively, you can use the fact that $\cos (90{}^{\circ }-\theta )=\sin \theta$ and $\sin (90{}^{\circ }-\theta )=\cos \theta$ and then write $\cos 60{}^{\circ }$ as $\cos (90{}^{\circ }-30{}^{\circ })=\sin 30{}^{\circ }$ and $\sin 60{}^{\circ }$ as $\sin (90{}^{\circ }-30{}^{\circ })=\cos 30{}^{\circ }$
Hence the expression becomes
$S={\sin }^{2}30{}^{\circ }+{\cos }^{2}30{}^{\circ }$
Now, we know that ${\sin }^2\theta +{\cos }^2\theta =1$
Hence, we have
$S=1$, which is the same as obtained above.
Hence option [c] is the correct answer.
[2] Alternative
We know that $\sin A\cos B+\cos A\sin B=\sin (A+B)$
Put $A=30{}^{\circ }$ and $B=60{}^{\circ }$.
Hence, we have
$\sin (30{}^{\circ }+60{}^{\circ })=\sin 30{}^{\circ }\cos 60{}^{\circ }+\cos 30{}^{\circ }\sin 60{}^{\circ }=S$
Hence, we have
$S=\sin 90{}^{\circ }$
From the above table, we have $\sin 90{}^{\circ }=1$
Hence, we have $S=1$, which is the same as obtained above.
Hence option [c] is the correct answer.