Trigonometric Functions Test 36

Given,

\(\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}\)

\(=2 \cos \frac{20^{\circ}+100^{\circ}}{2} \cos \frac{100^{\circ}-20^{\circ}}{2}+\cos 140^{\circ}\)\(\quad \quad (\because \cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2})\)

\(=2 \cos 60^{\circ} \cos 40^{\circ}+\cos 140^{\circ}\)

\(=2 \times \frac{1}{2} \cos 40^{\circ}+\cos 140^{\circ}\)\(\quad\quad(\because \cos 90^{\circ}=0)\)

\(=\cos 40^{\circ}+\cos 140^{\circ}\)

\(=2 \cos \frac{140^{\circ}+40^{\circ}}{2} \cos \frac{140^{\circ}-40^{\circ}}{2}\)\(\quad \quad (\because \cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2})\)

\(=2 \cos 90^{\circ} \cos 50^{\circ}\)\(\quad\quad(\because \cos 90^{\circ}=0)\)

\(=2 × 0×\cos 50^{\circ}\)

\(=0\)

Given,

\(\cos \theta+\sec \theta=\sqrt{3} \ldots(\mathrm{i})\)

\(\Rightarrow \cos \theta+\left(\frac{1}{\cos \theta}\right)=\sqrt{3} \ldots .\) (ii) \(\quad\left(\because \sec \theta=\frac{1}{\cos \theta}\right)\)

Cubing both side in equation (ii) we get,

\(\cos \theta+(\frac{1}{\cos \theta}^{3}=(\sqrt{3})^{3}\)

\(\Rightarrow \cos ^{3} \theta+(\frac{1}{\cos \theta})^3+\left\{3 \times \cos \theta \times\left(\frac{1}{\cos \theta}\right) \times\left[\cos \theta+\left(\frac{1}{\cos \theta} \right)\right]\right\}=3 \sqrt{3} \quad\left(\because(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right)\)

By using equation (ii) we get,

\(\Rightarrow \cos ^{3} \theta+\left(\frac{1}{\cos ^{3} \theta}\right)+3 \sqrt{3}=3 \sqrt{3}\)

\(\Rightarrow \cos ^{3} \theta+\left(\frac{1}{\cos ^{3} \theta}\right)=3 \sqrt{3}-3 \sqrt{3} \\\)

\(\Rightarrow \cos ^{3} \theta+\left(\frac{1}{\cos ^{3} \theta}\right)=0 \\\)

\(\Rightarrow \frac{\cos ^{6} \theta+1}{\cos ^{3} \theta}=0 \\\)

\(\Rightarrow \cos ^{6} \theta+1=0 \\\)

\(\Rightarrow \cos ^{6} \theta=(-1)\)

\(\sec ^{6} \theta=(-1)\)

\(\Rightarrow\left(\sec ^{2} \theta\right)^{3}=(-1)^3\)

\(\Rightarrow\left(\sec ^{2} \theta\right)=(-1) \quad\left(\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right)\)

\(\Rightarrow 1+\tan ^{2} \theta=-1\)

\(\Rightarrow \tan ^{2} \theta=-1-1 =(-2)\)

Value of \(\left(\cos ^{12} \theta+\tan ^{2} \theta\right):\left(\sec ^{6} \theta+\cos ^{6} \theta\right)=\left[(-1)^{2}+(-2)\right]:[(-1)+(-1)]\)

\(=(-1):(-2)\)

\(=1: 2\)

Given,

\(\sin A \sin \left(60^{\circ}-A\right) \sin \left(60^{\circ}+A\right)=k \sin 3 A\)

As we know that,

\(\sin (A+B)=\sin A \cos B+\cos A \sin B\) and \(\sin (A-B) =\sin A\) \(\cos B-\cos A\sin B \)

\(\Rightarrow \sin A \sin \left(60^{\circ}-A\right) \sin \left(60^{\circ}+A\right)=\sin A\left(\sin 60^{\circ} \cos A-\cos 60^{\circ}\sin A \right)\left(\sin 60^{\circ}\right.\) \(\left.\cos A+ \cos 60^{\circ}\sin A\right)\)

\(\Rightarrow\sin A[\frac{\sqrt{3}}{2}×\cos A-\frac{1}{2}×\sin A)(\frac{\sqrt{3}}{2}×\) \(\cos A+\frac{1}{2}×\sin A]\) = \(k\sin 3A\)

\(\Rightarrow\sin A[\frac{(\sqrt{3}×\cos A)^2-(\sin A)^2}{4}]\) = \(k\sin 3A\)

\(\Rightarrow\sin A[\frac{3×\cos ^2A-\sin ^2A}{4}]\) = \(k\sin 3A\)

\(\Rightarrow\sin A[\frac{3×(1-\sin ^2A)-\sin ^2A}{4}]\) = \(k\sin 3A\)\(\quad\quad (\because 1-\sin^2A = \cos^2 A)\)

\(\Rightarrow\sin A[\frac{(3-3\sin ^2A)-\sin ^2A}{4}]\) = \(k\sin 3A\)

\(\Rightarrow \frac{(3\sin A-4\sin ^3A)}{4}\) = \(k\sin 3A\)

\(\Rightarrow\frac{\sin 3A}{4}\) = \(k\sin 3A\)\(\quad\quad (\because \sin3A = 3\sin A-4\sin ^3A)\)

\(\Rightarrow k =\frac{1}{4}\)

Given,

\(\sec ^{2} \theta+\tan ^{2} \theta=7\) ......(i)

Adding 1 both sides in eqaution (i) we get,

\(\sec ^{2} \theta+\tan ^{2} \theta+1=7+1\)

\(\Rightarrow \sec ^{2} \theta+\sec ^{2} \theta=8\)\(\quad\quad(\because \ 1+\tan ^{2} \theta=\sec ^{2} \theta)\)

\(\Rightarrow 2 \sec ^{2} \theta=8\)

\(\Rightarrow \sec ^{2} \theta=4\)

\(\Rightarrow\sec \theta=2\)

\(\therefore \sec \theta=2\)

Given,

\(\sin x+\cos x=\sqrt{2}\)....(i)

By squaring both the sides in equation (i), we get

\((\sin x+\cos x)^2=(\sqrt{2})^2\)

\(\Rightarrow \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=(\sqrt{2})^{2}\)\(\quad\quad(\because (a +b)^2=a^{2} +b ^{2} +2 a b)\)

As we know that, \(\sin ^{2} x+\cos ^{2} x=1\) and \(\sin 2 x=2 \sin x \cos x\)

\(1+\sin 2 x=2\)

\(\Rightarrow \sin 2 x=1\)....(ii)

As we know that, \(\sin ^{2} x+\cos ^{2} x=1\)

Therefore,

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

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

By using equation (ii) we get,

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

\(\Rightarrow\cos ^{2} 2 x=0\)

Given:

\(\cos ^{2} \theta=\frac{(x+y)^{2}}{4 x y}\)

We know that,

Max value of \(\cos ^{2} \theta=1\)

Therefore,

\(1=\frac{(x+y)^{2}}{4 x y}\)

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

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

\(\Rightarrow 0=x^{2}+y^{2}-2 x y\)

\(\Rightarrow 0=(x-y)^{2}\)

\(\Rightarrow 0=x-y\)

\(\Rightarrow x=y\)

Given,

\(\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 359^{\circ}\)

We know that,

\(\sin (180-\theta)=\sin \theta\).....(i)

\(\sin (180+\theta)=-\sin \theta\)....(ii)

Therefore, Rewrite the given expression in terms of \((180-\theta)\) and \((180+\theta)\) we get,

\(\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin \left(180^{\circ}\right)+\sin \left(180^{\circ}+1^{\circ}\right)+\sin \left(180^{\circ}+2^{\circ}\right)+\cdots+\sin \left(180^{\circ}+179^{\circ}\right)\)

\(=\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin \left(179^{\circ}\right)+\sin \left(180^{\circ}\right)-\sin \left(1^{\circ}\right)-\sin \left(2^{\circ}\right)+\cdots-\sin \left(179^{\circ}\right)\)

\(=\sin 180^{\circ}\)

\(=0\)

Given,

\(\cot \left(22 \frac{1}{2}^{\circ}\right)\)

We know that, \(\cot x=\frac{\cos x}{\sin x}\)

\(\Rightarrow \cot \left(22 \frac{1}{2}^{\circ}\right)=\frac{\cos \left(22 \frac{1^{\circ}}{2}\right)}{\sin \left(22 \frac{1}{2}^{\circ}\right)}\)

\(\Rightarrow \sin \left(\frac{45^{\circ}}{2}\right)=\pm \sqrt{\frac{1-\cos \left(45^{\circ}\right)}{2}}\)\(\quad \quad (\because \sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}})\)

As we know that, \(0^{\circ}<\frac{\mathrm{A}}{2}<90^{\circ}\) where all trigonometric ratios are positive.

\(\Rightarrow \sin \left(\frac{45^{\circ}}{2}\right)=\sqrt{\frac{\sqrt{2}-1}{2 \sqrt{2}}}\)......(i)

\(\Rightarrow \cos \left(\frac{45^{\circ}}{2}\right)=\pm \sqrt{\frac{1+\cos \left(45^{\circ}\right)}{2}}\)\(\quad \quad (\because \cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}})\).....(ii)

As we know that, \(0^{\circ}<\frac{\mathrm{A}}{2}<90^{\circ}\) where all trigonometric ratio's are positive.

\(\Rightarrow \cos \left(\frac{45^{\circ}}{2}\right)=\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\)

So, from equation (i) and (ii), we get

\(\Rightarrow \cot \left(22 \frac{1}{2}^{\circ}\right)=\frac{\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}}{\sqrt{\frac{\sqrt{2}-1}{2 \sqrt{2}}}}\)

\(=1+\sqrt{2}\)

\(\Rightarrow \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}\)

Rationalization above the equation

\(\Rightarrow \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}}\)

\(\Rightarrow \sqrt{\frac{(\sqrt{2}+1)^{2}}{(\sqrt{2})^{2}-(1)^{2}}} \quad\left(\because a^{2}-b^{2}=(a+b)(a-b)\right.)\)

\(\Rightarrow \sqrt{\frac{(\sqrt{2}+1)^{2}}{2-1}}\)

\(=1+\sqrt{2}\)

Given,

\(\sin \theta+\sin 3 \theta+\sin 5 \theta+\operatorname{sin} 7 \theta\)

\(=(\sin 3 \theta+\sin \theta)+(\sin 7 \theta+\sin 5 \theta)\)

\(=2 \sin \left(\frac{3 \theta+\theta}{2}\right) \cos \left(\frac{3 \theta-\theta}{2}\right)+2 \sin \left(\frac{7 \theta+5 \theta}{2}\right) \cos \left(\frac{7 \theta-5 \theta}{2}\right)\)\(\quad\quad (\because\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right))\)

\(=(2 \sin 2 \theta \cos \theta)+(2 \sin 6 \theta \cos \theta)\)

\(=2 \cos \theta(\sin 2 \theta+\sin 6 \theta)\)

\(=2 \cos \theta(\sin 6 \theta+\sin 2 \theta)\)

\(=2 \cos \theta \times 2 \sin \left(\frac{6 \theta+2 \theta}{2}\right) \cos \left(\frac{6 \theta-2 \theta}{2}\right)\)\(\quad\quad (\because\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right))\)

\(=2 \cos \theta \times 2 \sin 4 \theta \cos 2 \theta\)

\(=4 \cos \theta \cos 2 \theta \sin 4 \theta\)

\(\therefore\) \(\sin \theta+\sin 3 \theta+\sin 5 \theta+\operatorname{sin} 7 \theta\) \(=4 \cos \theta \cos 2 \theta \sin 4 \theta\)

Given,

\(\sec ^{2} \theta+\tan ^{2} \theta=3\)...(i)

Subtracting 1 from both sides in equation (i) we get,

\(\sec ^{2} \theta+\tan ^{2} \theta-1=3-1\)\(\quad(\because\)\(\sec ^{2} \theta-1=\tan ^{2} \theta)\)

\(\Rightarrow \tan ^{2} \theta+\tan ^{2} \theta=2\)

\(\Rightarrow 2 \tan ^{2} \theta=2\)

\(\Rightarrow \tan ^{2} \theta=1\)

\(\Rightarrow \tan \theta=1\)

Now,

\(\cot \theta=\frac{1}{\tan \theta}\)

\(=1\)