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Question: How do you find the values of \[\sin 2\theta \] and \(\cos 2\theta \) when \(\cos \theta = \dfrac{{1...

How do you find the values of sin2θ\sin 2\theta and cos2θ\cos 2\theta when cosθ=1213?\cos \theta = \dfrac{{12}}{{13}}?

Explanation

Solution

From trigonometric ratios we know that cosine function is adjacent by hypotenuse. So we have the values of adjacent side and hypotenuse side, by using Pythagoras formula given by: Hypotenuse2=Adjacent2+Perpendicular2Hypotenus{e^2} = Adjacen{t^2} + Perpendicula{r^2} we can find the other side. By using trigonometric ratios we can find the required answer.

Complete step-by-step answer:
Now, we have given cosθ=1213\cos \theta = \dfrac{{12}}{{13}} , from the trigonometric ratios we know that cosine function is nothing but adjacenthypotenuse\dfrac{{adjacent}}{{hypotenuse}} which can be represented as in the below diagram

So in a triangle we know the values of adjacent and hypotenuse sides, the other side that is perpendicular or opposite can be find by using Pythagoras theorem formula given by Hypotenuse2=Adjacent2+Perpendicular2Hypotenus{e^2} = Adjacen{t^2} + Perpendicula{r^2}
Now, adjacent side BCBC value is 1212 and hypotenuse ACAC is 1313 let, xx be the perpendicular ABAB .
Therefore, we can write as AC2=BC2+AB2A{C^2} = B{C^2} + A{B^2}
132=122+x2\Rightarrow {13^2} = {12^2} + {x^2}
132122=x2\Rightarrow {13^2} - {12^2} = {x^2}
x2=169144\Rightarrow {x^2} = 169 - 144
x2=25\Rightarrow {x^2} = 25
x=5\Rightarrow x = 5
Therefore, the perpendicular value is 55 .
Now, the diagram is as follows:

By using trigonometric ratios we can write as follows:
Sine function is given by: oppositehypotenuse\dfrac{{opposite}}{{hypotenuse}} sinθ=513 \Rightarrow \sin \theta = \dfrac{5}{{13}}
Tangent function is given by: oppositeadjacent\dfrac{{opposite}}{{adjacent}} tanθ=512 \Rightarrow \tan \theta = \dfrac{5}{{12}}
In the question they have asked to find the values of sin2θ\sin 2\theta and cos2θ\cos 2\theta .
We have a formula for sin2θ\sin 2\theta which is given as sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta
By substituting the values in the above formula, we get
sin2θ=2×513×1213\sin 2\theta = 2 \times \dfrac{5}{{13}} \times \dfrac{{12}}{{13}}
On simplification, we get
sin2θ=120169\Rightarrow \sin 2\theta = \dfrac{{120}}{{169}}
To find cos2θ\cos 2\theta , we have a formula for cos2θ\cos 2\theta given by cos2θ=cos2θsin2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
By substituting the known values in the above cos2θ\cos 2\theta formula, we get
cos2θ=(1213)2(513)2\cos 2\theta = {\left( {\dfrac{{12}}{{13}}} \right)^2} - {\left( {\dfrac{5}{{13}}} \right)^2}
cos2θ=(144169)(25169)\Rightarrow \cos 2\theta = \left( {\dfrac{{144}}{{169}}} \right) - \left( {\dfrac{{25}}{{169}}} \right)
On simplification, we get
cos2θ=119169\Rightarrow \cos 2\theta = \dfrac{{119}}{{169}}
Therefore sin2θ=120169\sin 2\theta = \dfrac{{120}}{{169}} and cos2θ=119169\cos 2\theta = \dfrac{{119}}{{169}}.

Note: Whenever we have this type of problem we have to remember the trigonometric ratios for all the functions and also the double angle formulas which makes it easy to solve the problem. The double angle formula for cosine function, we can also use cos2θ=12sin2θ\cos 2\theta = 1 - 2{\sin ^2}\theta we get the same answer. But you need to remember the formulas.