Question

If cos theta = √3/2
show that 4 cos³theta - 3cos theta =0
​

Answer 1

Step-by-step explanation:

GIVEN:−

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

2

3

\bf{TO \: \: PROVE \colon - }TOPROVE:−

\bf{4 \cos ^{3} \theta} - 3 \cos \theta = - 14cos

3

θ−3cosθ=−1

SOLVING LHS

→ As we know

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

hypotenuse

prependicular

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

\begin{gathered}\bf{ \cos \theta \: = \frac{base}{hypotenuse}} \\ \\ \bf \implies \cos \theta = \frac{1}{2}\end{gathered}

cosθ=

hypotenuse

base

⟹cosθ=

2

1

now put the value and get the result

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

2

1

)

3

−3×

2

1

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

8

1

−

2

3

\begin{gathered}\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\end{gathered}

⟹

2

1

−

2

3

⟹

2

1−3

⟹

2

−2

⟹−1

\huge \sf{LHS = RHS}LHS=RHS

HENCE PROVED ★

Answer 2

Answer:

ya answer is zero.

Step-by-step explanation:

just put the value of cos theta in the given using RHL AND LHL.

Then verified LHL = RHL