Question

find the value of cos^-1(cos 4π/11)+sin^-1(sin 35π/11)​

Answer 1

Step-by-step explanation:

Formula :

• 1. (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx(x+y+z)

$\begin{gathered}\sf A = sin( \frac{2 \pi }{7} ) + sin( \frac{4 \pi }{7} ) + sin( \frac{8 \pi }{7} ) \\ \\ A^{2} = sin^{2}( \frac{2 \pi }{7} ) + sin^{2}( \frac{4 \pi }{7} ) + sin^{2}( \frac{8 \pi }{7} ) + 2sin( \frac{2 \pi }{7} )sin( \frac{4 \pi }{7} ) + 2sin( \frac{4 \pi }{7} )sin( \frac{8 \pi }{7} ) + 2sin( \frac{8 \pi }{7} )sin( \frac{2 \pi }{7} ) \\ \\B = cos( \frac{2 \pi }{7} ) + cos( \frac{4 \pi }{7} ) + cos( \frac{8 \pi }{7} )\end{gathered}$

$B^{2} = cos^{2}( \frac{2 \pi }{7} ) + cos^{2}( \frac{4 \pi }{7} ) + cos^{2}( \frac{8 \pi }{7} ) + 2cos( \frac{2 \pi }{7} )cos( \frac{4 \pi }{7} ) + 2cos( \frac{4 \pi }{7} )cos( \frac{8 \pi }{7} ) + 2cos( \frac{8 \pi }{7} )cos( \frac{2 \pi }{7}$

So,

$A^{2} + B^{2} = (sin^{2}\sf( \frac{2 \pi }{7} ) + cos^{2}( \frac{2 \pi }{7} )) + (sin^{2}( \frac{4 \pi }{7} ) + cos^{2}( \frac{4 \pi }{7} )) + (sin^{2}( \frac{8 \pi }{7} ) + cos^{2}( \frac{8 \pi }{7} )) + 2(sin( \frac{2 \pi }{7} )sin( \frac{4 \pi }{7} ) + cos( \frac{2 \pi }{7} )cos( \frac{4 \pi }{7} )) + 2(sin( \frac{4 \pi }{7} )sin( \frac{8 \pi }{7} ) + cos( \frac{4 \pi }{7} )cos( \frac{8 \pi }{7} )) + 2(sin( \frac{8 \pi }{7} )sin( \frac{2 \pi }{7} ) + cos( \frac{8 \pi }{7} )cos( \frac{2 \pi }{7} ))$

$\begin{gathered}\sf A^{2} + B^{2} = 1 + 1 + 1 + 2cos( \frac{2 \pi }{7} ) + 2cos( \frac{4 \pi }{7} ) + 2cos( \frac{6 \pi }{7} ) \\ \\ A^{2} + B^{2} = 3 + 2(cos( \frac{2 \pi }{7} ) + cos( \frac{6 \pi }{7} )) + 2cos( \frac{4 \pi }{7} ) \\ \\\sf A^{2} + B^{2} = 3 + 2(2cos( \frac{4 \pi }{7} )cos( \frac{2 \pi }{7} )) + 2cos( \frac{4 \pi }{7} )\end{gathered}$

$\begin{gathered}\sf A^{2} + B^{2} = 3 + 4cos( \frac{4 \pi }{7})cos( \frac{2 \pi }{7} ) + 2cos( \frac{4 \pi }{7} ) \\ \\\\ A^{2} + B^{2} = 3 + 2cos( \frac{4 \pi }{7})(2cos( \frac{2 \pi }{7} ) + 1)\\ \\\sqrt{A^{2} + B^{2}} =\sqrt{3 + 2cos( \frac{4 \pi }{7})(2cos( \frac{2 \pi }{7} ) + 1)}\end{gathered}$

Answer 2

Answer:

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