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Friday, July 24, 2020

tan15 degrees and sin 75 degrees

tan15 degrees and sin 75 degrees

11th cbse ncert trigonometry

5.Find the values of sin 75degrees and tan 15degrees

using trigonometry formula trigonometry identities

sin(A+B) =sinAcosB+cosAsinB


choose A=45degrees B=30degrees

sin(75degrees)
sin(45+30) =sin45cos30+cos45sin30 ( angles in degree )

=[1/(sqrt(2)][sqrt(3)/2] +[1/(sqrt(2)][1/2]

= {[sqrt(3) +1} / {2*sqrt(2)}

using trigonometry formula trigonometry identities 

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

choose A=45degrees B=30degrees

tan 15degrees

tan[45-30] = [tan45 -tan30]  / [1+tan45tan30] ( angles in degree )

=[1-{1/sqrt(3)}] / [1+(1){1/sqrt(3)}]

=[ sqrt(3) -1 ] / [sqrt(3)} + 1] introduce conjugate

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3) + 1] [ sqrt(3) -1 ] }

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3)}^2 - 1] }
 using identities (a-b)^2 and (a+b)(a-b)
={ [sqrt(3)]^2 - 2*sqrt(3) + 1 } / {3 - 1}

={3-2*sqrt(3) +1} / [2 ]

= [ 4-2*sqrt(3) ] / 2

={ 2 * [2- sqrt(3)]} / 2

= [2 - sqrt(3)]



3.3
5.Find the values of sin 75degrees and tan 15degrees
solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution
11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx
12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x
solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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