Trigonometric Functions and Identities
7.0 Trigonometric ratios of some useful angles
7.0 Trigonometric ratios of some useful angles
- sin${75^ \circ }$ = $\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$ = cos${15^ \circ }$
- cos${75^ \circ }$ = $\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$ = sin${15^ \circ }$
- tan${75^ \circ }$ = $2 + \sqrt 3 $ = cot${15^ \circ }$
- cot${75^ \circ }$ = $2 - \sqrt 3 $ = tan${15^ \circ }$
- sin${9^ \circ }$ = $\frac{{\sqrt {3 + \sqrt 5 }-\sqrt {5 - \sqrt 5 } }}{4}$ = cos${81^ \circ }$
- cos${9^ \circ }$ = $\frac{{\sqrt {3 + \sqrt 5 } + \sqrt {5 - \sqrt 5 } }}{4}$ = sin${81^ \circ }$
Question 2. Prove that $${\sec ^4}\theta - {\sec ^2}\theta = {\tan ^4}\theta + {\tan ^2}\theta $$
Solution: LHS: ${\sec ^4}\theta - {\sec ^2}\theta $ = ${\sec ^2}\theta ({\sec ^2}\theta - 1)$
= $(1 + {\tan ^2}\theta ){\tan ^2}\theta $
= ${\tan ^2}\theta + {\tan ^4}\theta $
= ${\tan ^4}\theta + {\tan ^2}\theta $
= R.H.S
L.H.S = R.H.S
Question 3. Prove that $$\frac{{\tan \left( {\frac{\pi }{4} + x} \right)}}{{\tan \left( {\frac{\pi }{4} - x} \right)}} = {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}$$
Solution: Let us solve the left-hand side of the given question using trigonometric function identities explained above.
L.H.S.: Applying the identity $7$ and $8$, we get $$\frac{{\tan \left( {\frac{\pi }{4} + x} \right)}}{{\tan \left( {\frac{\pi }{4} - x} \right)}} = \frac{{\frac{{\tan \frac{\pi }{4} + \tan x}}{{1 - \tan \frac{\pi }{4}.\tan x}}}}{{\frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}.\tan x}}}} = \frac{{\frac{{1 + \tan x}}{{1 - \tan x}}}}{{\frac{{1 - \tan x}}{{1 + \tan x}}}} = {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}$$ Therefore, L.H.S. = R.H.S.
Question 4. Prove that $$\cos 4x = 1 - 8{\sin ^2}x{\cos ^2}x$$
Solution: Let us solve the left-hand side of the given question using trigonometric function identities explained above.
L.H.S.: Applying the identity $11$, we get $$\cos 2(2x) = 1 - 2{\sin ^2}(2x)$$ Again apply identity $12$, we get $$\begin{equation} \begin{aligned} \cos 4x = 1 - 2{\left( {2\sin x\cos x} \right)^2} = 1 - 2\left( {4{{\sin }^2}x{{\cos }^2}x} \right) \\ \cos 4x = 1 - 8{\sin ^2}x{\cos ^2}x \\\end{aligned} \end{equation} $$ Therefore, L.H.S. = R.H.S.
Question 5. Find the value of $$\tan \frac{\pi }{8}$$
Solution: Let us assume $$x = \frac{\pi }{8}$$ Therefore, $$2x = \frac{\pi }{4}$$Now, applying the identity $13$, we get $$\begin{equation} \begin{aligned} \tan 2x = \frac{{2\tan x}}{{1 - {{\tan }^2}x}} \\ \tan \frac{\pi }{4} = \frac{{2\tan \frac{\pi }{8}}}{{1 - {{\tan }^2}\frac{\pi }{8}}} \\ 1 = \frac{{2\tan \frac{\pi }{8}}}{{1 - {{\tan }^2}\frac{\pi }{8}}} \\ \Rightarrow 1 - {\tan ^2}\frac{\pi }{8} = 2\tan \frac{\pi }{8} \\ \Rightarrow {\tan ^2}\frac{\pi }{8} + 2\tan \frac{\pi }{8} - 1 = 0 \\\end{aligned} \end{equation} $$ Put $\tan \frac{\pi }{8} = y$, we get $$\begin{equation} \begin{aligned} {y^2} + 2y - 1 = 0 \\ y = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4 \times 1 \times - 1} }}{{2 \times 1}} = - 1 \pm \sqrt 2 \\\end{aligned} \end{equation} $$ From here, we get $$\tan \frac{\pi }{8} = - 1 \pm \sqrt 2 $$ Since, $\frac{\pi }{8}$ lies in first quadrant and from our previous discussion we know that the value of $tan(x)$ in first quadrant is positive. Therefore, $$\tan \frac{\pi }{8} = \sqrt 2 - 1$$
