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```
Evaluate the value of (1+cosπ/8) (1+cos3π/8) (1+cos5π/8) (1+cos7π/8)

```
6 months ago

```							cos 5pi/8=cos(pi/2+pi/8)=-sin pi/8cos 3pi/8=cos(pi/2-pi/8)=sin pi/8cos 7pi/8=cos(pi-pi/8)=-cos pi/8. Now the expression (1+cos pi/8)(1+cos 3pi/8)(1+cos 5pi/8)(1+cos 7pi/8) can be written as,(1+cos pi/8)(1+sin pi/8)(1-sin pi/8)(1-cos pi/8)=[1-(cos pi/8)^2][1–(sin pi/8)^2]=(sin pi/8)^2×(cos pi/8)^2=1/4×(2 sin pi/8.cos pi/8)^2=1/4(sin 2x pi/8)^2=1/4(sin pi/4)^2=1/4×1/[(2)^1/2]^2=1/4×1/2=1/8(proven)
```
6 months ago Saurabh Koranglekar
8407 Points
``` ```
6 months ago
```							Dear student Hope your doubt gets cleared Please refer the solved examples of the below link https://www.askiitians.com/iit-jee-mathematics/trigonometry.aspx
```
6 months ago
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