Question

how do we solve theta in this?
sin(theta)cos(theta)=1/4

Answer 1

do you know that 2sin \(\theta\) cos\(\theta\)= sin 2\(\theta\)?

Answer 2

nope.

Answer 3

so, now you know it, then solve from then. Don't ask me "how" because I don't know.

Answer 4

how? haha. i don't know either.

Answer 5

I told you, don't ask me "how" hahaha.... I can predict that question. XD

Answer 6

Do you know sin(x+y) ?

Answer 7

@student12345 ?
sin2theta = 2.sintheta.cos theta!

Answer 8

how? @AkashdeepDeb can you do the procedure step by step? i feel confused.

Answer 9

Okay so there is one thing you'd HAVE to learn and that is
sin(x+y) = sinx.cosy + siny.cosx
Alright?

Answer 10

yes.

Answer 11

Now what is
\[\sin 2\theta = \sin (\theta + \theta) = \sin \theta.\cos \theta + \sin \theta.\cos \theta = 2\sin \theta.\cos \theta\]
According to the law above!
Getting this? :)

Answer 12

no. :s

Answer 13

sin(x+y) = sinx.cosy + siny.cosx
sin(θ+θ)=sinθ.cosθ+sinθ.cosθ=2sinθ.cosθ
x=y=\[\theta\]
Understood? :)

Answer 14

yes yes. but how could we rely it to sin(theta)cos(theta)=1/4 ?

Answer 15

Okay so now
Let us take it like this!
4.sinθ.cosθ = 1
Right? :)

Answer 16

or
2(2.sinθ.cosθ) = 1
or
sin2θ = 1/2 [Inverse of that previous statement]
So
sin2θ = sin 30 = 1/2
Or
2θ = 30
or
θ = 15'
Understood now? :D

Answer 17

we have formulae:
\[\sin(x+y) = sinx.cosy + cosx.siny\]
\[\sin(x-y) = sinx.cosy - cosx.siny\]
\[\cos(x+y) = cosx.cosy - sinx.siny\]
\[\cos(x-y) = cosx.cosy + sinx.siny\]
\[\tan(x+y) = \frac{ tanx + tany }{ 1 - tanx.tany }\]
\[\tan(x-y) = \frac{ tanx - tany }{ 1 + tanx.tany }\]
remember these

Answer 18

@student12345 can u now give me 2θ's of the three trig functions, cos , sin and tan from above equations?

Answer 19

the 2θ's are important for u , u need to by-heart them after you learned how they actually derived.