![geometry - there is any relation between $\pi$, $\sqrt{2}$ or a generic polygon? - Mathematics Stack Exchange geometry - there is any relation between $\pi$, $\sqrt{2}$ or a generic polygon? - Mathematics Stack Exchange](https://i.stack.imgur.com/yPEWc.jpg)
geometry - there is any relation between $\pi$, $\sqrt{2}$ or a generic polygon? - Mathematics Stack Exchange
![SOLVED: For Sin a = 2/3 and Cos b = -1/4 with both in [pi/2.pi) Find Cos ( a-b) [2 Sart(15) - Sqrt(5) ] / 12 [2 Sqrt (5) Sqrt (3) ]/ 12 SOLVED: For Sin a = 2/3 and Cos b = -1/4 with both in [pi/2.pi) Find Cos ( a-b) [2 Sart(15) - Sqrt(5) ] / 12 [2 Sqrt (5) Sqrt (3) ]/ 12](https://cdn.numerade.com/ask_images/4150e8e2965a46589e6825184fa18722.jpg)
SOLVED: For Sin a = 2/3 and Cos b = -1/4 with both in [pi/2.pi) Find Cos ( a-b) [2 Sart(15) - Sqrt(5) ] / 12 [2 Sqrt (5) Sqrt (3) ]/ 12
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Amazon.com: Funny 25 Years Old Pi Math Square Root Of 625| 25th Birthday Long Sleeve T-Shirt : Clothing, Shoes & Jewelry
![Ifint(-pi/4)^((3pi)/4)(e^(pi/4)dx)/((e^x+e^(pi/4))(sinx+cosx))=kint(-pi/2)^( pi/2)secx dx , then the value of k is (a) 1/2 (b) 1/(sqrt(2)) (c) 1/(2sqrt(2)) (d) -1/(sqrt(2)) Ifint(-pi/4)^((3pi)/4)(e^(pi/4)dx)/((e^x+e^(pi/4))(sinx+cosx))=kint(-pi/2)^( pi/2)secx dx , then the value of k is (a) 1/2 (b) 1/(sqrt(2)) (c) 1/(2sqrt(2)) (d) -1/(sqrt(2))](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/642535683_web.png)