If sin(2A)= cos75°,find the smallest positive value of A
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Hey !!! ^_^
Here is your answer
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Sin2A = Cos75° .....{SinA = Cos(90 - A)}
Cos(90°- 2A) = Cos75°
Removing Cos From LHS and RHS
90° - 2A = 75°
2A = 90° - 75°
2A = 15
A = 15/2
A = 7.5
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I HOPE IT WILL HELP YOU
Thank you
☺️
Here is your answer
⬇️⬇️⬇️⬇️⬇️⬇️
Sin2A = Cos75° .....{SinA = Cos(90 - A)}
Cos(90°- 2A) = Cos75°
Removing Cos From LHS and RHS
90° - 2A = 75°
2A = 90° - 75°
2A = 15
A = 15/2
A = 7.5
=====================÷=
I HOPE IT WILL HELP YOU
Thank you
☺️
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