Question

The value of (1+cos pi/10) (1+cos 3pi/10) (1+cos 7pi/10) (1+cos 9pi/10)

Answer 1

Answer: 1/16

Step-by-step explanation:

We know,

$1)sin(\pi/10)=\frac{(\sqrt{5}-1)}{4}\\ \\ 2)sin(3\pi/10)=\frac{(\sqrt{5}+1)}{4} \\ \\ 3) (1-cos^2(\theta)) = sin^2(\theta) \\ \\ 4) cos(\pi-\theta) =-cos(\theta)$

Now, moving swiftly to the given equation.

$(1+cos(\pi/10)(1+cos(3\pi/10)(1+cos(7\pi/10)(1+cos(9\pi/10) \\ \\ = (1+cos(\pi/10)(1-cos(7\pi/10)(1+cos(7\pi/10)(1-cos(\pi/10) \\ \\=(1-cos^2(\pi/10)(1-cos^2(7\pi/10) \\ \\=sin^2(\pi/10)sin^2(7\pi/10) \\ \\=[(\frac{(\sqrt{5}-1}{4}) (\frac{(\sqrt{5}-1}{4})]^2\\ \\=[\frac{5-1}{4*4}]^2 \\ \\=\frac{1}{16}$

Hence, Required value is 1/16.

Answer 2

Answer:

1/16

Step-by-step explanation:

)sin(π/10)=

4

(

5

−1)

2)sin(3π/10)=

4

(

5

+1)

3)(1−cos

2

(θ))=sin

2

(θ)

4)cos(π−θ)=−cos(θ)

Now, moving swiftly to the given equation.

\begin{lgathered}(1+cos(\pi/10)(1+cos(3\pi/10)(1+cos(7\pi/10)(1+cos(9\pi/10) \\ \\ = (1+cos(\pi/10)(1-cos(7\pi/10)(1+cos(7\pi/10)(1-cos(\pi/10) \\ \\=(1-cos^2(\pi/10)(1-cos^2(7\pi/10) \\ \\=sin^2(\pi/10)sin^2(7\pi/10) \\ \\=[(\frac{(\sqrt{5}-1}{4}) (\frac{(\sqrt{5}-1}{4})]^2\\ \\=[\frac{5-1}{4*4}]^2 \\ \\=\frac{1}{16}\end{lgathered}

(1+cos(π/10)(1+cos(3π/10)(1+cos(7π/10)(1+cos(9π/10)

=(1+cos(π/10)(1−cos(7π/10)(1+cos(7π/10)(1−cos(π/10)

=(1−cos

2

(π/10)(1−cos

2

(7π/10)

=sin

2

(π/10)sin

2

(7π/10)

=[(

4

(

5

−1

)(

4

(

5

−1

)]

2

=[

4∗4

5−1

]

2

=

16

1