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त्रिकोणमितीय फलन
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NCERT HINDI-त्रिकोणमितीय फलन-प्रश्नावली 3.3
- cos .((pi)/(4)-x) cos ((pi)/(4)-y) –sin ((pi)/(4) –x) sin ((pi)/(4)-y)...
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- (tan((pi)/(4)+x))/(tan((pi)/(4)-x))=((1+tanx)/(1-tan x))^(2)
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- (cos (pi+x) cos (-x))/(sin (pi-x) cos ((pi)/(2)+x))=cot^(2)x
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- cos ((3pi)/(2)+x)cos (2pi+x)[cot ((3pi)/(2)-x)+cot (2pi+x)]=1
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- sin(n+1)x sin(n+2)x+ cos (n+1)x cos (n+2)x=cos x
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- cos((3pi)/(4)+x)-cos ((3pi)/(4)-x)=-sqrt(2) sin x
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- sin^(2)6x-sin^(2)4x=sin2x sin 10x
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- cos^(2) 2x - cos^(2) 6x = sin 4 x sin 8x
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- sin2 x+2 sin 4x + sin 6x=4 cos^(2) x sin 4x
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- cot 4x (sin 5x + sin 3x)=cot x (sin 5x-sin 3x)
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- (cos 9x-cos 5x)/(sin 17x-sin 3x)=-(sin2x)/(cos 10x)
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- (sin 5x+ sin 3x)/(cos 5x+ cos 3x)=tan 4x
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- (sin x- sin y)/(cos x+ cos y)=tan ""(x-y)/(2)
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- (sin x+sin 3x)/(cos x+ cos 3x)=tan 2x
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- (sin x- sin 3x)/(sin^(2) x- cos^(2)x)=2 sin x
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- (cos 4x+ cos 3x+ cos2x)/(sin 4x+ sin 3x+ sin 2x)=cot 3x
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- cot x cot 2x -cot 2x cot 3x -cot 3x cot x=1
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- tan 4x=(4tan x(1-tan^(2)x))/(1-6tan^(2)x+tan^(4)x)
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- cos 4x=1-8sin^(2) x cos^(2)x
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- cos 6x = 32 cos^(6) x-48 cos^(4) x +18 cos^(2) x-1
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