If θ = 30°, verify that
Question:

If $\theta=30^{\circ}$, verify that

(i) $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$

(ii) $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

(iii) $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$

(iv) $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$

Solution:

(i) Given:’

$\theta=30^{\circ}$….(1)

To verify:

$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$….(2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

LHS $=\tan 2 \theta$

Now by substituting the value of $\theta$ from equation (1) in the above expression

We get,

$\mathrm{LHS}=\tan 2 \times\left(30^{\circ}\right)$

$=\tan 60^{\circ}$

$=\sqrt{3}$

Now by substituting the value of $\theta$ from equation (1) in the expression $\frac{2 \tan \theta}{1-\tan ^{2} \theta}$

We get,

$\mathrm{RHS}=\frac{2 \tan \left(30^{\circ}\right)}{1-\tan ^{2}\left(30^{\circ}\right)}$….(4)

$\mathrm{RHS}=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}$

$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1^{2}}{(\sqrt{3})^{2}}}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}$

$=\sqrt{3}$

Now by comparing equation (3) and (4)

We get,

$\mathrm{LHS}=\mathrm{RHS}=\sqrt{3}$

Hence $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}$

(ii) Given:

$\theta=30^{\circ} \ldots \ldots(1)$

To verify:

$\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$…(2)

$\sin 2 \theta=\sin 2 \times 30$

$=\sin 60$

 

$=\frac{\sqrt{3}}{2}$

Now consider right hand side

$\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\frac{2 \tan 30}{1+\tan ^{2} 30}$

$=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$

$=\frac{\sqrt{3}}{2}$

Hence it is verified that,

$\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

(iii) Given:

$\theta=30^{\circ}$….(1)

To verify:

$\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$….(2)

Now consider left hand side of the equation (2)

Therefore,

$\cos 2 \theta=\cos 2 \times 30$

$=\cos 60$

$=\frac{1}{2}$

Now consider right hand side of equation (2)

Therefore,

$\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{1-(\tan 30)^{2}}{1+(\tan 30)^{2}}$

$=\frac{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$

$=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}$

$=\frac{1}{2}$

Hence it is verified that,

$\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$

(iv) Given:

$\theta=30^{\circ}$…(1)

To verify:

$\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \ldots \ldots(2)$

Now consider left hand side of the expression in equation (2)

Therefore

$\cos 3 \theta=\cos 3 \times 30$

$=\cos 90$

 

$=0$

Now consider right hand side of the expression to be verified in equation (2)

Therefore,

$4 \cos ^{3} \theta-3 \cos \theta=4 \cos ^{3} 30-3 \cos 30$

$=4 \times\left(\frac{\sqrt{3}}{2}\right)^{3}-3 \times \frac{\sqrt{3}}{2}$

$=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}$

$=0$

Hence it is verified that,

$\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$

 

 

 

 

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