Question

Show that \[\sin {{100}^{\circ }}-\sin {{10}^{\circ }}\] is positive.

Answer 1

Hint: In this problem, we have to show that \[\sin {{100}^{\circ }}-\sin {{10}^{\circ }}\] is positive. We can first multiply and divide \[\sqrt{2}\] in the given expression. We can then take the common term separately outside the bracket. We can then convert the value into trigonometric form inside the bracket. We can now use the formula \[\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)\] to find whether the given expression is positive or not. We will get the answer in cosine form, which is positive in the first quadrant.

Complete step by step solution:
We know that the given expression is,
\[\sin {{100}^{\circ }}-\sin {{10}^{\circ }}\]
We can now multiply and divide \[\sqrt{2}\] in the above expression, we get
\[\Rightarrow \dfrac{\sqrt{2}}{\sqrt{2}}\sin {{100}^{\circ }}-\dfrac{\sqrt{2}}{\sqrt{2}}\sin {{10}^{\circ }}\]
We can now take the common term separately.
\[\Rightarrow \sqrt{2}\left( \dfrac{1}{\sqrt{2}}\sin {{100}^{\circ }}-\dfrac{1}{\sqrt{2}}\sin {{10}^{\circ }} \right)\] …….. (1)
We know that,
\[\Rightarrow \cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}=\sin {{45}^{\circ }}\] ……… (2)
We also know that,
\[\Rightarrow \sin {{100}^{\circ }}=\sin \left( {{90}^{\circ }}+{{10}^{\circ }} \right)=\cos {{10}^{\circ }}\] ……… (3)
We can now substitute (2) and (3) in (1), we get
\[\Rightarrow \sqrt{2}\left( \cos {{45}^{\circ }}\cos {{10}^{\circ }}-\sin {{45}^{\circ }}\sin {{10}^{\circ }} \right)\]……. (4)
Here we have substituted \[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}=\sin {{45}^{\circ }}\] according to the formula form.
We can see that (4) is of the form.
\[\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)\]
We can now write (4) as,
\[\begin{align}
  & \Rightarrow \sqrt{2}\cos \left( {{45}^{\circ }}+{{10}^{\circ }} \right) \\
 & \Rightarrow \sqrt{2}\cos {{55}^{\circ }} \\
\end{align}\]
Which is positive, since cos is positive in the first quadrant.

Therefore, \[\sin {{100}^{\circ }}-\sin {{10}^{\circ }}=\sqrt{2}\cos {{55}^{\circ }}\]which is positive.

Note: Students make mistakes while substituting \[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}=\sin {{45}^{\circ }}\]in (1), as we have to get a form which should be similar to trigonometric identity. We should also remember some trigonometric formulas to solve these types of problems. We should also know that cos is always positive in the first quadrant. We should remember the trigonometric formula \[\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)\].