Trigonometric Functions Test 37

Given,

\(\sin \left(-1125^{\circ}\right)\)

\(=-\sin \left(1125^{\circ}\right)\) \(\quad\quad (\because\) \(\sin (-x)=-\sin x)\)

\(=-\sin \left(360^{\circ} \times 3+45^{\circ}\right)\)\(\quad\quad(\because\) \(\sin (2 n \pi+x)=\sin x)\)

\(=-\sin 45^{\circ}\)

\(=\frac{-1}{\sqrt{2}}\)

\(\therefore\) \(\sin \left(-1125^{\circ}\right)\) \(=\frac{-1}{\sqrt{2}}\).

Given,

\(\sin x+\sin 3 x+\sin 5 x=0\)

\(\Rightarrow(\sin 5 \mathrm{x}+\sin \mathrm{x})+\sin 3 \mathrm{x}=0\)

\(\Rightarrow 2 \sin 3 x \cos 2 x+\sin 3 x=0\) \(\quad (\because \sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2})\)

\(\Rightarrow \sin 3 x(2 \cos 2 x+1)=0\)

\(\sin 3 \mathrm{x}=0\) or \(2 \cos 2 \mathrm{x}+1=0\)

\(\sin 3 \mathrm{x}=0\) or, \(\cos 2 \mathrm{x}=-\frac{1}{2}\)

Now, \(\sin 3 \mathrm{x}=0 \Longrightarrow 3 \mathrm{x}=\mathrm{n} \pi \Longrightarrow \mathrm{x}=\frac{\mathrm{n} \pi}{3}, \mathrm{n} \in \mathrm{Z}\)

And, \(\cos 2 \mathrm{x}=-\frac{1}{2}\)

\(\cos 2 \mathrm{x}=\cos \frac{2 \pi}{3}\)

\(2 \mathrm{x}=2 \mathrm{~m} \pi \pm \frac{2 \pi}{3}, \mathrm{~m} \in \mathrm{Z}\)

\(\mathrm{x}=\mathrm{m} \pi \pm \frac{\pi}{3}, \mathrm{~m} \in \mathrm{Z}\)

\(\therefore\) The general solution of the given equation is : \(\mathrm{x}=\frac{\mathrm{n} \pi}{3}\) or, \(\mathrm{x}=\mathrm{m} \pi \pm \frac{\pi}{3}\), where \(\mathrm{m}, \mathrm{n} \in \mathrm{Z}\).

Given,

\(\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\)...(i)

We know that,

\(\operatorname{sin}(A+B)=(\operatorname{sin} A \times \cos B+\operatorname{cos} A \times \operatorname{sin} B)\)....(ii)

Comparing equation (i) with euation (ii) we get,

\(A=25^{\circ}\)

\(B=65^{\circ}\)

\(\operatorname{sin}\left(25^{\circ}+65^{\circ}\right)\)

\(\Rightarrow \operatorname{sin} 90^{\circ}=1\)

Put, \(\theta=60^{\circ}\)

\(\Rightarrow \cos \theta>\cos ^{2} \theta\)

\(\Rightarrow \cos 60^{\circ}>\cos ^{2} 60^{\circ}\)

\(\Rightarrow \frac{1}{2}>\frac{1}{4}\) \(\cos \theta>\cos ^{2} \theta\)

Given,

\(\sin x+\operatorname{sin}^{2} x=1\)

\(\Rightarrow \sin x=1-\sin ^{2} x\) \(\quad\quad (\because 1 - \sin ^{2} \theta =\cos ^{2} \theta)\)

\(\Rightarrow \sin x=\cos ^{2} x \quad\) ......(i)

Sqauring both in equation (i) we get,

\(\sin ^{2} x=\cos ^{4} x\)

\(\Rightarrow 1-\cos ^{2} x=\cos ^{4} x \) \(\quad\quad (\because \sin ^{2} \theta =1 - \cos ^{2} \theta)\)

\(\Rightarrow \cos ^{4} x+\cos ^{2} x=1\)

Given,

\(\cos \left(3015^{\circ}\right)\)

\(=\cos \left(360^{\circ} \times 8^{\circ}+135^{\circ}\right)\)\(\quad\quad(\because\cos (2 n \pi \pm \theta)=\cos \theta)\)

\(=\cos \left(135^{\circ}\right)\)

\(=\cos \left(90^{\circ}+45^{\circ}\right) \)\(\quad\quad(\because\cos (90+\theta)=-\sin \theta)\)

\(=-\sin 45^{\circ}\)

\(=\frac{-1}{\sqrt{2}}\)

Given,

\(\sin x+\sin (x-\pi)+\sin (x+\pi)\)

We know that,

\(\sin (\pi-x)=\sin x\)

\(\sin (\pi+x)=-\sin x\)

\(\sin (-x)=-\sin x \)

Therefore,

\(=\sin x-\sin (\pi-x)-\sin (\pi+x)\)

\(=\sin x-\sin (\pi-x)-\sin x\)

\(=-\sin (\pi-x)\)

\(=-\sin x\)

Given,

\(1+\cot ^{2} \theta\)

\(=1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\)\(\quad\quad(\because\cot\theta =\frac{\cos\theta}{\\sin\theta})\)

\(=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta}\) \(\quad\quad(\because\sin ^{2} \theta+\cos ^{2} \theta=1)\)

\(=\frac{1}{\sin ^{2} \theta}\)

\(=\operatorname{cosec}^{2} \theta\)

Given,

\(\cos \left(-1230^{\circ}\right) \times \cot \left(-315^{\circ}\right)\)

\(\cos \left(-1230^{\circ}\right)=\cos \left(1230^{\circ}\right)\)\(\quad\quad(\because \cos (-\theta)=\cos \theta)\)

\(=\cos \left(3 \times 360^{\circ}+150\right)\)\(\quad\quad(\because\cot (2 n \pi-\theta)=-\cot (\theta))\)

\(=\cos \left(150^{\circ}\right)\)

\(=\cos \left(180^{\circ}-30^{\circ}\right)\)\(=-\cos 30^{\circ} \quad(\because \cos (\pi-\theta)=-\cos (\theta))\)

\(=\frac{-\sqrt{3}}{2}\)

Now,

\(\cot \left(-315^{\circ}\right)=-\cot \left(315^{\circ}\right) \quad(\because \cot (-\theta)=-\cot (\theta))\)

\(=-\cot \left(360^{\circ}-45^{\circ}\right) \)

\(=\cot 45^{\circ} \quad(\because \cot (2 n \pi-\theta)=-\cot (\theta))\)

\(=1\)

\(\cos \left(-1230^{\circ}\right) \times \cot \left(-315^{\circ}\right)=\frac{-\sqrt{3}}{2}\)

\(=1 × \frac{-\sqrt{3}}{2}\)

Given,

\(\frac{\sin 7 x+6 \sin 5 x+17 \sin 3 x+12 \sin x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=\frac{\sin 7 \mathrm{x}+\sin 5 \mathrm{x}+5 \sin 5 \mathrm{x}+5 \sin 3 \mathrm{x}+12 \sin 3 \mathrm{x}+12 \sin \mathrm{x}}{\sin 6 \mathrm{x}+5 \sin 4 \mathrm{x}+12 \sin 2 \mathrm{x}}\)

By using, \(\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\)

Therefore,

\(=\frac{(2 \sin(\frac{7x+5x}{2}) \cos (\frac{7x-5x}{2})+5×(2\sin(\frac{5x+3x}{2})\cos (\frac{5x-3x}{2}))+12×(2\sin(\frac{3x+x}{2}) \cos(\frac{3x-x}{2}))}{(\sin 6 x+5 \sin 4 x+12 \sin 2 x)}\)

\(=\frac{2 \sin 6 x \cos x+10 \sin 4 x \cos x+24 \sin 2 x \cos x}{(\sin 6 x+5 \sin 4 x+12 \sin 2 x)}\)

\(=\frac{2 \cos x(\sin 6 x+5 \sin 4 x+12 \sin 2 x)}{(\sin 6 x+5 \sin 4 x+12 \sin 2 x)}\)

\(=2 \cos x\)