Question

sin theta-cos theta=√6-√2/2, then tha value of 4(sin^3-cos^3)^2 is equal to​

Answer 1

$\textbf{Given:}$

$\mathsf{sin\theta-cos\theta=\dfrac{\sqrt{6}-\sqrt{2}}{2}}$

$\textbf{To find:}$

$\textsf{The value of}\;\mathsf{4(sin^3\theta-cos^3\theta)^2}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{sin\theta-cos\theta=\dfrac{\sqrt{6}-\sqrt{2}}{2}}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{(sin\theta-cos\theta)^2=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2}\right)^2}$

$\mathsf{sin^2\theta+cos^2\theta-2\,sin\theta\,cos\theta=\dfrac{6+2-2\sqrt{12}}{4}}$

$\mathsf{1-2\,sin\theta\,cos\theta=\dfrac{8-4\sqrt{3}}{4}}$

$\mathsf{1-2\,sin\theta\,cos\theta=2-\sqrt{3}}$

$\mathsf{1-2+\sqrt{3}=2\,sin\theta\,cos\theta}$

$\mathsf{-1+\sqrt{3}=2\,sin\theta\,cos\theta}$

$\implies\mathsf{sin\theta\,cos\theta=\dfrac{\sqrt{3}-1}{2}}$.........(1)

$\textsf{Now}$

$\mathsf{sin^3\theta-cos^3\theta}$

$\mathsf{=(sin\theta-cos\theta)(sin^2\theta+sin\theta\,cos\theta+cos^2\theta)}$

$\mathsf{=(sin\theta-cos\theta)(1+sin\theta\,cos\theta)}$

$\mathsf{=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2}\right)\left(1+\dfrac{\sqrt{3}-1}{2}\right)}$

$\mathsf{=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}+1}{2}\right)}$

$\mathsf{=\left(\dfrac{\sqrt{2}(\sqrt{3}-1)}{2}\right)\left(\dfrac{\sqrt{3}+1}{2}\right)}$

$\mathsf{=\dfrac{\sqrt{2}(\sqrt{3}^2-1^2)}{4}}$

$\mathsf{=\dfrac{\sqrt{2}(3-1)}{4}}$

$\mathsf{=\dfrac{2\sqrt{2}}{4}}$

$\mathsf{=\dfrac{\sqrt{2}}{2}}$

$\mathsf{Now,}$

$\mathsf{4(sin^3\theta-cos^3\theta)^2}$

$\mathsf{=4\left(\dfrac{\sqrt{2}}{2}\right)^2}$

$\mathsf{=4\left(\dfrac{2}{4}\right)}$

$\mathsf{=2}$

$\implies\boxed{\mathsf{4(sin^3\theta-cos^3\theta)^2=2}}$

$\textbf{Find more:}$

If 3x=cosecθ and 3/x=cotθ, then find the value of 3(x²-1/x²) PLEASE HELP BEST ANSWER WILL BE MARKED BRAINLIEST

https://brainly.in/question/21134289

Answer 2

Answer:

sinθ−cosθ=

2

6

−

2

\textbf{To find:}To find:

\textsf{The value of}\;\mathsf{4(sin^3\theta-cos^3\theta)^2}The value of4(sin

3

θ−cos

3

θ)

2

\textbf{Solution:}Solution:

\mathsf{Consider,}Consider,

\mathsf{sin\theta-cos\theta=\dfrac{\sqrt{6}-\sqrt{2}}{2}}sinθ−cosθ=

2

6

−

2

\textsf{Squaring on bothsides, we get}Squaring on bothsides, we get

\mathsf{(sin\theta-cos\theta)^2=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2}\right)^2}(sinθ−cosθ)

2

=(

2

6

−

2

)

2

\mathsf{sin^2\theta+cos^2\theta-2\,sin\theta\,cos\theta=\dfrac{6+2-2\sqrt{12}}{4}}sin

2

θ+cos

2

θ−2sinθcosθ=

4

6+2−2

12

\mathsf{1-2\,sin\theta\,cos\theta=\dfrac{8-4\sqrt{3}}{4}}1−2sinθcosθ=

4

8−4

3

\mathsf{1-2\,sin\theta\,cos\theta=2-\sqrt{3}}1−2sinθcosθ=2−

3

\mathsf{1-2+\sqrt{3}=2\,sin\theta\,cos\theta}1−2+

3

=2sinθcosθ

\mathsf{-1+\sqrt{3}=2\,sin\theta\,cos\theta}−1+

3

=2sinθcosθ

\implies\mathsf{sin\theta\,cos\theta=\dfrac{\sqrt{3}-1}{2}}⟹sinθcosθ=

2

3

−1

.........(1)

\textsf{Now}Now

\mathsf{sin^3\theta-cos^3\theta}sin

3

θ−cos

3

θ

\mathsf{=(sin\theta-cos\theta)(sin^2\theta+sin\theta\,cos\theta+cos^2\theta)}=(sinθ−cosθ)(sin

2

θ+sinθcosθ+cos

2

θ)

\mathsf{=(sin\theta-cos\theta)(1+sin\theta\,cos\theta)}=(sinθ−cosθ)(1+sinθcosθ)

\mathsf{=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2}\right)\left(1+\dfrac{\sqrt{3}-1}{2}\right)}=(

2

6

−

2

)(1+

2

3

−1

)

\mathsf{=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}+1}{2}\right)}=(

2

6

−

2

)(

2

3

+1

)

\mathsf{=\left(\dfrac{\sqrt{2}(\sqrt{3}-1)}{2}\right)\left(\dfrac{\sqrt{3}+1}{2}\right)}=(

2

2

(

3

−1)

)(

2

3

+1

)

\mathsf{=\dfrac{\sqrt{2}(\sqrt{3}^2-1^2)}{4}}=

4

2

(

3

2

−1

2

)

\mathsf{=\dfrac{\sqrt{2}(3-1)}{4}}=

4

2

(3−1)

\mathsf{=\dfrac{2\sqrt{2}}{4}}=

4

2

2

\mathsf{=\dfrac{\sqrt{2}}{2}}=

2

2

\mathsf{Now,}Now,

\mathsf{4(sin^3\theta-cos^3\theta)^2}4(sin

3

θ−cos

3

θ)

2

\mathsf{=4\left(\dfrac{\sqrt{2}}{2}\right)^2}=4(

2

2

)

2

\mathsf{=4\left(\dfrac{2}{4}\right)}=4(

4

2

)

\mathsf{=2}=2

\implies\boxed{\mathsf{4(sin^3\theta-cos^3\theta)^2=2}}⟹

4(sin

3

θ−cos

3

θ)

2

=2

\textbf{Find more:}Find more:

If 3x=cosecθ and 3/x=cotθ, then find the value of 3(x²-1/x²)