The correct option is D sin2θ Have we see an expression similar to 1 2sin^2(π/4 + θ) before? Yes, cos2x = 1 2sin^2x Here x = π/4 + θ ⇒ 1 2sin^2(π/4 + θ ...

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Trigonometric Ratios for Sum of Two Angles

1-2 sin 2π/4+...

Question

1 - 2 $s i n^{2} (\frac{π}{4} + θ)$ =

cos2θ

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-cos2θ

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sin2θ

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-sin2θ

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Solution

The correct option is D
-sin2θ

Have we see an expression similar to 1 - 2 $s i n^{2} (\frac{π}{4} + θ)$ before?

Yes,

cos2x = 1 - $2 s i n^{2} x$

Here x = $\frac{π}{4} + θ$

$\Rightarrow$ 1 - 2 $s i n^{2} (\frac{π}{4} + θ)$ = $c o s 2 (\frac{π}{4} + θ)$

= $c o s (\frac{π}{4} + 2 θ)$

= $- s i n^{2} θ$

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Similar questions

Q. Identity transformations of Trigonometric Expressions.
prove the following identities.

\frac{2 s i n^{2} (\frac{π}{4} - α)}{c o s 2 α} = c o t (\frac{π}{4} + α) .

2 s i n^{2} ((\frac{π}{2}) c o s^{2} x) = 1 - c o s (π s i n 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

, then

c o s 2 x

is equal to

Q. If

{cos}^{4} θ + {cos}^{2} θ = 1

then show that
i)

{sec}^{4} θ - {sec}^{2} θ = 1

ii)

{cot}^{4} θ - {cot}^{2} θ = 1

Q. if

2 {sin}^{2} ((\frac{π}{2}) {cos}^{2} x) = 1 - cos (π {sin}^{2} 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

,then

cos 2 x

is equal to

Q. The value of

θ

laying between

θ = 0

and

θ = π / 2

and satisfying the equation

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

are

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1 - 2sin2(π4+θ) =

The correct option is D -sin2θ Have we see an expression similar to 1 - 2sin2(π4+θ) before? Yes, cos2x = 1 - 2sin2x Here x = π4+θ ⇒ 1 - 2sin2(π4+θ) = cos2(π4+θ) = cos(π4+2θ) = −sin2θ

1 - 2 $s i n^{2} (\frac{π}{4} + θ)$ =