Trigonometry | Part -3

Trigonometric Identities and Applications

i) Show that if \( \cos\theta = \frac{1}{\sqrt{2}} \), then \( 3\sin\theta - 4\sin^3\theta = \frac{1}{\sqrt{2}} \)

From the identity: \( \sin^2\theta + \cos^2\theta = 1 \)
\( \Rightarrow \sin^2\theta = 1 - \cos^2\theta = 1 - \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \Rightarrow \sin\theta = \pm \frac{1}{\sqrt{2}} \)

Let's take \( \sin\theta = \frac{1}{\sqrt{2}} \) (same sign as \( \cos\theta \), assuming \( \theta \) is in the first quadrant):
\[ 3\sin\theta - 4\sin^3\theta = 3\left(\frac{1}{\sqrt{2}}\right) - 4\left(\frac{1}{\sqrt{2}}\right)^3 = \frac{3}{\sqrt{2}} - \frac{4}{2\sqrt{2}} = \frac{3}{\sqrt{2}} - \frac{2}{\sqrt{2}} = \frac{1}{\sqrt{2}} \] Hence, verified.

ii) If \( \cot\theta = \frac{4}{3} \), evaluate \( \frac{3\cos\theta + 5\sin\theta}{6\cos\theta - 2\sin\theta} \)

Given: \( \cot\theta = \frac{4}{3} \Rightarrow \frac{\cos\theta}{\sin\theta} = \frac{4}{3} \)
So, assume \( \cos\theta = 4k \), \( \sin\theta = 3k \) for some constant \( k \).

Using the identity: \( \cos^2\theta + \sin^2\theta = 1 \)
\[ (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2 = 1 \Rightarrow k = \frac{1}{5} \]

Now: \( \cos\theta = \frac{4}{5} \), \( \sin\theta = \frac{3}{5} \)

Substituting in the expression: \[ \frac{3\cos\theta + 5\sin\theta}{6\cos\theta - 2\sin\theta} = \frac{3 \times \frac{4}{5} + 5 \times \frac{3}{5}}{6 \times \frac{4}{5} - 2 \times \frac{3}{5}} = \frac{\frac{12 + 15}{5}}{\frac{24 - 6}{5}} = \frac{\frac{27}{5}}{\frac{18}{5}} = \frac{27}{18} = \frac{3}{2} \]

Final Answer: \( \frac{3}{2} \)

iii) In triangle ABC, right-angled at B, if AB = 5 cm and BC + AC = 25 cm, find the value of \( \sin^2 A + \cos^2 A \)

For any angle \( A \), we have the identity:
\[ \sin^2 A + \cos^2 A = 1 \] So, the answer is simply:
\[ \sin^2 A + \cos^2 A = 1 \] regardless of the triangle dimensions, as long as angle A is defined.