Question

if sinA=3/5then find the values of the following sin 3A,cos3A,tan3A,cot3A
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Answer 1

Answer:

sinA=3/5.

we have ,

sin^2 A+cos^2 A=1

=(3/5)^2+cos^2 A=1

=9/25+cos^2 A=1

=cos^2 A=1-9/25

=cos^2 A=16/25

=cosA=4/5.

tan A=sin A/cos A

=3/5÷4/5

=3/4

cos 3A=4cos^3 A- 3cos A

=4×(4/5)^3 - 3(4/5)

=4×64/125 - 12/5

=256/125 - 300/125

=44/125

=0.352

tan 3A=(3 tan A-tan^3 A)/(1-3tan^2 A)

=(3×3/4-(3/4)^3)/(1-3 (3/4)^2)

=(9/4-27/64)/(1-27/16)

=(144-27/64)/(16-27/16)

=117/64×16/-11

=117/4×-1/11

=(-117/44)

=2.65

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Answer 2

Answer:

दो क्रमागत सम संख्याओ का H.C.F. ज्ञात कीजिए