Question

Find the value of trigonometric ratios given as:
\[1)\sin {{30}^{\circ }}\cos {{30}^{\circ }}\]
\[2)\tan {{30}^{\circ }}\tan {{60}^{\circ }}\]
\[3){{\cos }^{2}}{{60}^{\circ }}+{{\sin }^{2}}30\]

Answer 1

Hint: To solve this type of problem we have to know the exact values of trigonometric functions. Write the value of trigonometric function and do mathematical operations like multiplication for the first to and addition for the last one.

Complete step-by-step solution -
\[1)\sin {{30}^{\circ }}\cos {{30}^{\circ }}\]
The value of \[\sin {{30}^{\circ }}=\dfrac{1}{2}\] . . . . . . . . . . . . . . . . . . . . (a)
The value of \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]. . . . . . . . . . . . . . . . . . . . (b)
Now substituting (a) and (b) in \[\sin {{30}^{\circ }}\cos {{30}^{\circ }}=\left( \dfrac{1}{2} \right)\times \left( \dfrac{\sqrt{3}}{2} \right)\]
\[\sin {{30}^{\circ }}\cos {{30}^{\circ }}\]\[=\dfrac{\sqrt{3}}{4}\]
\[2)\tan {{30}^{\circ }}\tan {{60}^{\circ }}\]
The value of \[\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
The value of \[\tan {{60}^{\circ }}=\sqrt{3}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now substituting (1) and (2) in \[\tan {{30}^{\circ }}\tan {{60}^{\circ }}=\left( \dfrac{1}{\sqrt{3}} \right)\times \left( \sqrt{3} \right)\]
\[\tan {{30}^{\circ }}\tan {{60}^{\circ }}\]= \[=1\]
\[3){{\cos }^{2}}{{60}^{\circ }}+{{\sin }^{2}}30\]
The value of \[\cos {{60}^{\circ }}=\dfrac{1}{2}\]
The value of \[{{\cos }^{2}}{{60}^{\circ }}=\dfrac{1}{4}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . (p)
The value of \[\sin {{30}^{\circ }}=\dfrac{1}{2}\]
The value of \[{{\sin }^{2}}{{30}^{\circ }}=\dfrac{1}{4}\]. . . . . . . . . . . . . . . . . . . . . . (q)
Now substituting (p) and (q) in \[{{\cos }^{2}}{{60}^{\circ }}+{{\sin }^{2}}30\] we get
\[{{\cos }^{2}}{{60}^{\circ }}+{{\sin }^{2}}30=\dfrac{1}{4}+\dfrac{1}{4}\]
\[{{\cos }^{2}}{{60}^{\circ }}+{{\sin }^{2}}30=\dfrac{1}{2}\].

Note: From (s) we may think it as trigonometric identity but it is not because in trigonometric identity both the trigonometric functions have the same value of \[\theta \].This is an easy and direct problem. Take care during calculations. Here we can also use the trigonometric identities by doing some conversions.