Question

The value of $\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$ is _______.

Answer 1

Hint: Since the trigonometric function consists of $\sin {\theta}$ and $\cos {\theta}$, use the sine-cosine trigonometric relations to simplify the given function.
i.e. $\text{sin}\left( \text{A+B} \right)\text{=sinA}\text{.cosB + cosA}\text{.sinB}$
Then substitute the standard values of given angles in the equation to get an answer.

Complete step-by-step solution:
In the given question, we have:
$\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$
By using identity, $\text{sin}\left( \text{A+B} \right)\text{=sinA}\text{.cosB + cosA}\text{.sinB}$
We can write equation (1) as:
$\begin{align}
  & \text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}=\text{ sin}\left( \text{3}{{0}^{\circ }}+{{60}^{\circ }} \right) \\
 & =\text{ sin}{{90}^{\circ }}
\end{align}$
Since $\text{sin}{{90}^{\circ }}=1$;
Therefore,
$\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}=\text{ 1}$
Hence the value of $\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$ is 1.

Note: This is another way to calculate the given trigonometric function. We know that the expression contains all standard angles, so we can directly put the values of standard angles in the equation and solve it to get the answer, i.e.
$\text{sin}{{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\text{sin3}{{0}^{\circ }}\text{= cos}{{60}^{\circ }}=\dfrac{1}{2}$
The table below shows the values of standard angles:

\[\angle A\]	${{0}^{\circ }}$	${{30}^{\circ }}$	${{45}^{\circ }}$	${{60}^{\circ }}$	${{90}^{\circ }}$
$\sin A$	0	$\dfrac{1}{2}$	\[\dfrac{1}{\sqrt{2}}\]	$\dfrac{\sqrt{3}}{2}$	1
$\cos A$	1	$\dfrac{\sqrt{3}}{2}$	\[\dfrac{1}{\sqrt{2}}\]	$\dfrac{1}{2}$	0
$\tan A$	0	\[\dfrac{1}{\sqrt{3}}\]	1	$\sqrt{3}$	Not defined
$\text{cosec }A$	Not defined	2	\[\sqrt{2}\]	\[\dfrac{2}{\sqrt{3}}\]	1
$\sec A$	1	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	Not defined
$\cot A$	Not defined	$\sqrt{3}$	1	\[\dfrac{1}{\sqrt{3}}\]	0

Therefore, using the above values and substituting in the given expression, we get:
$\begin{align}
  & \text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}=\left( \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} \right)+\left( \dfrac{1}{2}\times \dfrac{1}{2} \right) \\
 & =\dfrac{3}{4}+\dfrac{1}{4} \\
 & =\dfrac{4}{4} \\
 & =1
\end{align}$
Yet, we can solve the given expression by using more complex identities:
$\left\{ \begin{align}
  & 2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right) \\
 & 2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right) \\
\end{align} \right\}$
In the given question, we have:
$\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$
By multiplying and dividing the whole equation by 2, we get:
\[\Rightarrow \dfrac{1}{2}\left( \text{2sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ 2sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }} \right)\]
Now, by applying the identities stated above, we can write the expression as:
\[\Rightarrow \dfrac{1}{2}\left( \begin{align}
  & \left( \text{sin}\left( {{60}^{\circ }}+{{30}^{\circ }} \right)+\text{sin}\left( {{60}^{\circ }}-{{30}^{\circ }} \right) \right) \\
 & +\left( \text{sin}\left( {{60}^{\circ }}+{{30}^{\circ }} \right)-\text{sin}\left( {{60}^{\circ }}-{{30}^{\circ }} \right) \right) \\
\end{align} \right)\]
Now, solving the equations, we get:
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}\left( 2\text{sin}\left( {{60}^{\circ }}+{{30}^{\circ }} \right) \right) \\
 & \Rightarrow \text{sin}\left( {{60}^{\circ }}+{{30}^{\circ }} \right) \\
 & \Rightarrow \sin {{90}^{\circ }} \\
 & \Rightarrow 1 \\
\end{align}\]
Hence, we get the same result, i.e. the value of $\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$ is 1.
One more way to solve this expression is to convert sine into cosine and vice-versa, i.e.
For the given expression: $\text{sin}{{60}^{\circ }}\cos {{30}^{\circ }}+\text{ sin3}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$
We can convert \[\cos {{30}^{\circ }}\Rightarrow \sin \left( {{90}^{\circ }}-{{30}^{\circ }} \right)\] and $\sin {{30}^{\circ }}\Rightarrow \cos \left( {{90}^{\circ }}-{{30}^{\circ }} \right)$
So, we get:
$\Rightarrow \text{sin}{{60}^{\circ }}\sin {{60}^{\circ }}+\text{ cos6}{{0}^{\circ }}\text{ cos}{{60}^{\circ }}$
We can solve this expression putting the values for standard angles. So, we get:
$\begin{align}
  & \Rightarrow \left( \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} \right)+\left( \dfrac{1}{2}\times \dfrac{1}{2} \right) \\
 & \Rightarrow \dfrac{3}{4}+\dfrac{1}{4} \\
 & \Rightarrow 1 \\
\end{align}$
Hence, we can say that whichever method we try, we are getting the same value. Therefore, there are various methods to solve a trigonometric equation.