Question

How do you find the exact values of $\sin {{35}^{\circ }}\cos {{60}^{\circ }}-\cos {{35}^{\circ }}\sin {{60}^{\circ }}$ using the sum and difference, double angle or half angle formulas?

Answer 1

Hint: From the question given it had been asked to find the exact value of $\sin {{35}^{\circ }}\cos {{60}^{\circ }}-\cos {{35}^{\circ }}\sin {{60}^{\circ }}$. By using the sum and difference, double angle or half angle formulas. We can use the difference formula for sine to solve the above given question.
The difference formula for sine is given below: $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$

Complete step by step answer:
Now considering from the question we have the trigonometric expression \[\sin {{35}^{\circ }}\cos {{60}^{\circ }}-\cos {{35}^{\circ }}\sin {{60}^{\circ }}\] .
From the basic concepts we know that we have a similar formula given as \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\] .
By using this formula we will have a simplified trigonometric expression given as \[\sin {{35}^{\circ }}\cos {{60}^{\circ }}-\cos {{35}^{\circ }}\sin {{60}^{\circ }}=\sin \left( {{35}^{\circ }}-{{60}^{\circ }} \right)\] .
By further simplifying this expression we will have $\sin \left( -{{25}^{\circ }} \right)$ .
As we know that the value of $\sin \left( -\theta \right)$ is given as $-\sin \theta $.
Therefore we have $-\sin \left( {{25}^{\circ }} \right)$ .
We now have to find the exact value of $-\sin \left( {{25}^{\circ }} \right)$ .
In trigonometric we have one formula given as $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta $
By substituting $\theta ={{25}^{\circ }}$ we will have \[\sin {{75}^{\circ }}=3\sin {{25}^{\circ }}-4{{\sin }^{3}}{{25}^{\circ }}\] .
As we know that
\[\begin{align}
  & \sin {{45}^{\circ }}\cos {{30}^{\circ }}-\cos {{45}^{\circ }}\sin {{30}^{\circ }}=\sin \left( {{75}^{\circ }} \right) \\
 & \Rightarrow \dfrac{1}{\sqrt{2}}\left( \dfrac{\sqrt{3}}{2} \right)-\dfrac{1}{\sqrt{2}}\left( \dfrac{1}{2} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}} \\
\end{align}\]
By substituting the value of $\sin \left( {{75}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ we will have
\[\begin{align}
  & \sin {{75}^{\circ }}=3\sin {{25}^{\circ }}-4{{\sin }^{3}}{{25}^{\circ }} \\
 & \Rightarrow \dfrac{\sqrt{3}-1}{2\sqrt{2}}=3\sin {{25}^{\circ }}-4{{\sin }^{3}}{{25}^{\circ }} \\
\end{align}\]
By simplifying $\sin \left( {{75}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$ we will have $\dfrac{1.732-1}{2\left( 1.414 \right)}=\dfrac{0.732}{2.828}=\dfrac{732}{2828}=0.258$ .
By substituting $x=\sin {{25}^{\circ }}$ and $\sin {{75}^{\circ }}=0.258$ we will have \[0.258=3x-4{{x}^{3}}\] .
After solving this cubic equation $4{{x}^{3}}-3x+0.258=0$ we will have $x=0.422$ .
As we have $\sin {{25}^{\circ }}=0.422$ .

Therefore we can conclude that
$\sin 35\cos 60-\cos 35\sin 60=-\sin \left( 25 \right)$
$\Rightarrow \sin 35\cos 60-\cos 35\sin 60=-0.42262$.

Note: We should be well aware of the trigonometric angles. We should also be well aware of the formulas of sum and difference, and half angle and double angle formulas. We should use the basic trigonometric formula to get the exact values of the angles. Calculation must be done very carefully by us while doing the solution of the given question. Similarly we have formulae like $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$, $\cos \left( a-b \right)=\sin a\sin b+\cos a\cos b$ and $\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$ .