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Question: Find \(\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1\). A. \(2\sin 2\theta \) ...

Find sin⁡3θ−cos⁡3θsin⁡θ+cos⁡θ+1\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1.
A. 2sin⁡2θ2\sin 2\theta
B. 2cos⁡2θ2\cos 2\theta
C. tan⁡2θ\tan 2\theta
D. cot⁡2θ\cot 2\theta

Explanation

Solution

We first use multiple angle formulas and complete the summation with 1. We use the identity formulas like sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta . Then we take 4(sin⁡θ+cos⁡θ)4\left( \sin \theta +\cos \theta \right) common to cancel it out from the denominator.

Complete answer:
We have the trigonometric multiple angle formula where sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta and cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta .
We place these values in sin⁡3θ−cos⁡3θsin⁡θ+cos⁡θ+1\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1 and simplify
sin⁡3θ−cos⁡3θsin⁡θ+cos⁡θ+1 =3sin⁡θ−4sin⁡3θ−4cos⁡3θ+3cos⁡θsin⁡θ+cos⁡θ+1 =4sin⁡θ−4sin⁡3θ−4cos⁡3θ+4cos⁡θsin⁡θ+cos⁡θ \begin{aligned} & \dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1 \\\ & =\dfrac{3\sin \theta -4{{\sin }^{3}}\theta -4{{\cos }^{3}}\theta +3\cos \theta }{\sin \theta +\cos \theta }+1 \\\ & =\dfrac{4\sin \theta -4{{\sin }^{3}}\theta -4{{\cos }^{3}}\theta +4\cos \theta }{\sin \theta +\cos \theta } \\\ \end{aligned}
Now we try to take 4 common and factorise the numerator.
4sin⁡θ−4sin⁡3θ−4cos⁡3θ+4cos⁡θ =4(sin⁡θ+cos⁡θ)−4(sin⁡3θ+cos⁡3θ) \begin{aligned} & 4\sin \theta -4{{\sin }^{3}}\theta -4{{\cos }^{3}}\theta +4\cos \theta \\\ & =4\left( \sin \theta +\cos \theta \right)-4\left( {{\sin }^{3}}\theta +{{\cos }^{3}}\theta \right) \\\ \end{aligned}
We use the cubic expansion as a3+b3=(a+b)(a2+b2−ab){{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right).
So, we get sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(sin⁡2θ+cos⁡2θ−sin⁡θcos⁡θ){{\sin }^{3}}\theta +{{\cos }^{3}}\theta =\left( \sin \theta +\cos \theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right).
We know the identity value of sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.
So, sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(1−sin⁡θcos⁡θ){{\sin }^{3}}\theta +{{\cos }^{3}}\theta =\left( \sin \theta +\cos \theta \right)\left( 1-\sin \theta \cos \theta \right).
We now again take common of (sin⁡θ+cos⁡θ)\left( \sin \theta +\cos \theta \right).
4(sin⁡θ+cos⁡θ)−4(sin⁡3θ+cos⁡3θ) =4(sin⁡θ+cos⁡θ)−4(sin⁡θ+cos⁡θ)(1−sin⁡θcos⁡θ) =4(sin⁡θ+cos⁡θ)(1−1+sin⁡θcos⁡θ) =4sin⁡θcos⁡θ(sin⁡θ+cos⁡θ) \begin{aligned} & 4\left( \sin \theta +\cos \theta \right)-4\left( {{\sin }^{3}}\theta +{{\cos }^{3}}\theta \right) \\\ & =4\left( \sin \theta +\cos \theta \right)-4\left( \sin \theta +\cos \theta \right)\left( 1-\sin \theta \cos \theta \right) \\\ & =4\left( \sin \theta +\cos \theta \right)\left( 1-1+\sin \theta \cos \theta \right) \\\ & =4\sin \theta \cos \theta \left( \sin \theta +\cos \theta \right) \\\ \end{aligned}
The fraction form becomes sin⁡3θ−cos⁡3θsin⁡θ+cos⁡θ+1=4sin⁡θcos⁡θ(sin⁡θ+cos⁡θ)(sin⁡θ+cos⁡θ)=4sin⁡θcos⁡θ\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1=\dfrac{4\sin \theta \cos \theta \left( \sin \theta +\cos \theta \right)}{\left( \sin \theta +\cos \theta \right)}=4\sin \theta \cos \theta .
We know the identity formula of 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta .
So, 4sin⁡θcos⁡θ=2(2sin⁡θcos⁡θ)=2sin⁡2θ4\sin \theta \cos \theta =2\left( 2\sin \theta \cos \theta \right)=2\sin 2\theta .
Final solution is sin⁡3θ−cos⁡3θsin⁡θ+cos⁡θ+1=2sin⁡2θ\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1=2\sin 2\theta .
And hence the correct answer is option A.

Note:
It is important to remember that the condition to eliminate the (sin⁡θ+cos⁡θ)\left( \sin \theta +\cos \theta \right) from both denominator and numerator is (sin⁡θ+cos⁡θ)≠0\left( \sin \theta +\cos \theta \right)\ne 0. No domain is given for the variable xx. The value of tan⁡x≠−1\tan x\ne -1 is essential. The simplified condition will be x≠nπ−π4,n∈Zx\ne n\pi -\dfrac{\pi }{4},n\in \mathbb{Z}.