Question

If sin theta = under root 3 / 2, show that 4 cos 3 theta - 3 cos theta = - 1.

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

$\bf{GIVEN \colon - }$

$\bf{sin \theta \: = \frac{ \sqrt{3} }{2}}$

$\bf{TO \: \: PROVE \colon - }$

$\bf{4 \cos ^{3} \theta} - 3 \cos \theta = - 1$

SOLVING LHS

→ As we know

$\bf{ sin \theta \: = \frac{prependicular}{hypotenuse}}$

Hence

→ perpendicular = √3 units

→ hypotenuse = 2 units

so , let's find base

According to Pythagoreas theorem

» (hypotenuse)² = (perpendicular)² + (base)²

Hence

» (base)² = (perpendicular)² -(hypotenuse)²

» (base)² = (2)² - (√3)²

» (base)² = 4 - 3

→ base = √1

→ base = 1 unit

Hence

$\bf{ \cos \theta \: = \frac{base}{hypotenuse}} \\ \\ \bf \implies \cos \theta = \frac{1}{2}$

now put the value and get the result

$\bf \implies \: 4 \times (\frac{1}{2} ) ^{3} - 3 \times \frac{1}{2}$

$\bf \implies \: 4 \times \frac{1}{8} - \frac{3}{2}$

$\huge \bf \implies \frac{1}{2} - \frac{3}{2} \\ \\ \huge \bf \implies \: \frac{ 1 - 3}{2} \\ \\ \huge \bf \implies \: \frac{ - 2}{2} \\ \\ \huge \tt \implies \: - 1$

$\huge \sf{LHS = RHS}$

HENCE PROVED ★

Answer 2

Answer:

Step-by-step explanation: