Trigonometry

From Example Problems
Jump to: navigation, search

Trigonometry

  • \sin ^{2}(u)+\cos ^{2}(u)=1\,
  • e^{{i\theta }}=\cos(\theta )+i\sin(\theta )\,

  • {\mathrm  {Real}}\left[\exp(\pm i\theta )\right]=\cos(\theta )\,

  • \sin(x)=\sum _{{k=0}}^{\infty }{\frac  {(-1)^{k}x^{{2k+1}}}{(2k+1)!}}\,

  • \cos(x)=\sum _{{k=0}}^{\infty }{\frac  {(-1)^{k}x^{{2k}}}{(2k)!}}\,


TRIGONOMETRY BOOKS


solution Invert the matrix {\begin{bmatrix}\cos &\sin \\-\sin &\cos \\\end{bmatrix}}

solution Find the amplitude and period of 4\sin({\frac  {1}{3}}x)\,

solution If \sin(x)={\frac  {-4}{5}}\, and x\, is in the third quadrant, then find \cos(x)\,.

solution If a 10 foot tall ladder leans against a wall, and the base of the ladder is 5 feet away from the wall, then how far up the wall does the ladder go?

solution If an 11 foot tall ladder leans against a wall, and the base of the ladder makes a 24 degree angle with the ground, then how far up the wall does the ladder go?

solution If a 30 foot tall ladder leans against a wall, and the base of the ladder makes a 65 degree angle with the ground, then how far up the wall does the ladder go?



Basics

solution If \cos \theta +\sin \theta ={\sqrt  {2}}\cos \theta \, prove that \cos \theta -\sin \theta ={\sqrt  {2}}\sin \theta \,

solution Show that {\frac  {1+\sin A-\cos A}{1+\sin A+\cos A}}+{\frac  {1+\sin A+\cos A}{1+\sin A-\cos A}}=2\csc A\,

solution Prove that \cot ^{2}\theta [{\frac  {\sec \theta -1}{1+\sin \theta }}]+\sec ^{2}\theta [{\frac  {\sin \theta -1}{1+\sec \theta }}]=0\,

solution If \tan \theta +\sin \theta =m,\tan \theta -\sin \theta =n\, show that m^{2}-n^{2}=4{\sqrt  {mn}}\,

solution Eliminate\theta \, from a\cos \theta +b\sin \theta +c=0\, and a_{1}\cos \theta +b_{1}\sin \theta +c_{1}=0\,

solution If {\frac  {x}{a}}\sin \theta +{\frac  {y}{b}}\cos \theta =1\, and {\frac  {x}{a}}\cos \theta -{\frac  {y}{b}}\sin \theta =1\, show that {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=2\,

solution Prove that {\frac  {\csc A}{\csc A-1}}+{\frac  {\csc A}{\csc A+1}}=2\sec ^{2}A\,

solution Simplify {\sqrt  {{\frac  {1+\sin A}{1-\sin A}}}}=\sec A+\tan A\,

solution Prove that \sin ^{6}\theta +\cos ^{6}\theta =1-3\sin ^{2}\theta \cdot \cos ^{2}\theta \,

solution Simplify {\frac  {\sin ^{3}\theta +\cos ^{3}\theta }{\sin \theta +\cos \theta }}\,

solution Prove that {\frac  {\sin ^{4}\theta -\cos ^{4}\theta }{\sin ^{2}\theta -\cos ^{2}\theta }}=1\,

solution Prove that {\frac  {\tan ^{3}\theta -1}{\tan \theta -1}}=\sec ^{2}\theta +\tan \theta \,

solution Show that {\frac  {1+\sin \theta }{\cos \theta }}+{\frac  {\cos \theta }{1+\sin \theta }}=2\sec \theta \,

solution Prove that {\frac  {\tan A}{\sec A-1}}+{\frac  {\tan A}{\sec A+1}}=2\csc A\,

TRIGONOMETRY BOOKS


Compound Angles

\sin(A\pm B)=\sin A\cos B\pm \cos A\sin B\,

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

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

\tan(A\pm B)={\frac  {\tan A\pm \tan B}{1\mp \tan A\tan B}}\,

\cot(A+B)={\frac  {\cot A\cot B-1}{\cot B+\cot A}}\,

\cot(A-B)={\frac  {\cot A\cot B+1}{\cot B-\cot A}}\,



solution Prove that \sin(A+B)\sin(A-B)=\sin ^{2}A-\sin ^{2}B\, and \cos(A+B)\cos(A-B)=\cos ^{2}A-\sin ^{2}B,\,

solution Find the value of \tan[{\frac  {\pi }{4}}+A]\,

solution Find the value of \cos 105^{\circ },\sin 75^{\circ }\,

solution what is the value of {\frac  {\tan 40+\tan 20}{\cot 45-\cot 50\cot 70}}\,

solution Show that \cos 40+\cos 80+\cos 160=0\,

solution Prove that \tan 50=\tan 40+2\tan 10\,

solution Prove that \cos ^{2}A+\cos ^{2}B-2\cos A\cos B\cos(A+B)=\sin ^{2}(A+B)\,

solution Prove that \sin ^{2}\theta +\sin ^{2}(\theta +60)+\sin ^{2}(\theta -60)={\frac  {3}{2}}\,

solution IF {\frac  {m+1}{m-1}}={\frac  {\cos(\alpha -\beta )}{\sin(\alpha +\beta )}}\, then prove that m=\tan[{\frac  {\pi }{4}}+\alpha ]\tan[{\frac  {\pi }{4}}+\beta ]\,

solution In a triangle ABC if \cot A+\cot B+\cot C={\sqrt  {3}}\, then show that the triangle is equilateral.

solution A+B+C=180^{\circ }\,, prove that \tan A+\tan B+\tan C=\tan A\tan B\tan C\,

solution If \tan \beta ={\frac  {n\tan \alpha }{1+(1-n)\tan ^{2}\alpha }}\, then show that \tan(\alpha -\beta )=(1-n)\tan \alpha \,

solution If A+B=45, prove that (1+\tan A)(1+\tan B)=2\,.Hence show that \tan {\frac  {45}{2}}={\sqrt  {2}}-1\,

solution Prove that \tan(A-B)+\tan(B-C)+\tan(C-A)=\tan(A-B)\tan(B-C)\tan(C-A)\,

solution Prove that\tan(\theta -{\frac  {3\pi }{4}})\tan({\frac  {7\pi }{4}}+\theta )+1=0\,

solution Show that \cos ^{2}\theta +\cos ^{2}(60+\theta )+\cos ^{2}(60-\theta )={\frac  {3}{2}}\,

solution Show that \cos A+\cos(240-A)+\cos(240+A)=0\,

Multiple and Submultiple angles

1. \sin 2A=2\sin A\cos A,\sin A=2\sin {\frac  {A}{2}}\cos {\frac  {A}{2}}\,

2. \sin 2A={\frac  {2\tan A}{1+\tan ^{2}A}},\sin A={\frac  {2\tan {\frac  {A}{2}}}{1+\tan ^{2}{\frac  {A}{2}}}}\,

3. \sin 3A=3\sin A-4\sin ^{3}A\,

4. \cos 2A=\cos ^{2}A-\sin ^{2}A=2\cos ^{2}A-1=1-2\sin ^{2}A\,

5. \cos A=\cos ^{2}{\frac  {A}{2}}-\sin ^{2}{\frac  {A}{2}}=2\cos ^{2}{\frac  {A}{2}}-1=1-2\sin ^{2}{\frac  {A}{2}}\,

6. \cos 2A={\frac  {1-\tan ^{2}A}{1+\tan ^{2}A}},\cos A={\frac  {1-\tan ^{2}{\frac  {A}{2}}}{1+\tan ^{2}{\frac  {A}{2}}}}\,

TRIGONOMETRY BOOKS


7. \cos 3A=4\cos ^{3}A-3\cos A\,

8. \tan 2A={\frac  {2\tan ^{2}A}{1-\tan ^{2}A}},\tan A={\frac  {2\tan {\frac  {A}{2}}}{1-\tan ^{2}{\frac  {A}{2}}}}\,

9. \tan 3A={\frac  {3\tan A-\tan ^{3}A}{1-3\tan ^{2}A}}\,

solution Prove that {\frac  {\cos 3A+\sin 3A}{\cos A-\sin A}}=1+2\sin 2A\,

solution Show that \cos ^{6}A-\sin ^{6}A=\cos 2A[1-{\frac  {\sin ^{2}2A}{4}}]\,

solution Prove that \cot({\frac  {\pi }{4}}-\theta )={\frac  {\cos 2\theta }{1-\sin 2\theta }}\,. Hence find the value of \cot 15^{\circ }\,

solutionIf \tan A={\frac  {1-\cos B}{\sin B}}\,,then prove that \tan 2A=\tan B\,

solution Prove that \cos({\frac  {\pi }{11}})\cos({\frac  {2\pi }{11}})\cos({\frac  {3\pi }{11}})\cos({\frac  {4\pi }{11}})\cos({\frac  {5\pi }{11}})={\frac  {1}{32}}\,

solution Prove that [1+\cos {\frac  {\pi }{8}}][1+\cos {\frac  {3\pi }{8}}][1+\cos {\frac  {5\pi }{8}}][1+\cos {\frac  {7\pi }{8}}]={\frac  {1}{8}}\,

solution Prove that \sin A\sin[{\frac  {\pi }{3}}+A]\sin[{\frac  {\pi }{3}}-A]={\frac  {1}{4}}\sin 3A\,.Hence show that \sin {\frac  {\pi }{9}}\sin {\frac  {2\pi }{9}}\sin {\frac  {3\pi }{9}}\sin {\frac  {4\pi }{9}}={\frac  {3}{16}}\,

solution Prove that 16\cos ^{5}\theta -20\cos ^{3}\theta +5\cos \theta =\cos 5\theta \,

solution If m\tan(\theta -30)=n\tan(\theta +120)\, show that \cos 2\theta ={\frac  {m+n}{2(m-n)}}\,

solution Prove that \sin ^{4}{\frac  {\pi }{8}}+\sin ^{4}{\frac  {3\pi }{8}}+\sin ^{4}{\frac  {5\pi }{8}}+\sin ^{4}{\frac  {7\pi }{8}}={\frac  {3}{2}}\,

solution Prove that \cot \theta +\cot(60+\theta )-\cot(60-\theta )=3\cot 3\theta \,

Transformations

For all C,D\in R\,

1. \sin C+\sin D=2\sin {\frac  {C+D}{2}}\cos {\frac  {C-D}{2}}\,

2. \sin C-\sin D=2\cos {\frac  {C+D}{2}}\sin {\frac  {C-D}{2}}\,

3. \cos C+\cos D=2\cos {\frac  {C+D}{2}}\cos {\frac  {C-D}{2}}\,

4. \cos C-\cos D=-2\sin {\frac  {C+D}{2}}\sin {\frac  {C-D}{2}}\,

5. 2\sin A\cos B=\sin(A+B)+\sin(A-B)\,

6. 2\cos A\sin B=\sin(A+B)-\sin(A-B)\,

7. 2\cos A\cos B=\cos(A+B)+\cos(A-B)\,

8.-2\sin A\sin B=\cos(A+B)-\cos(A-B)\,




TRIGONOMETRY BOOKS












Trigonometric Equations

Inverse Trigonometric Functions

find the value of  sec x . cosx + sin ^2 ( x ) + cos ^2( x )

Hyperbolic Functions

Properties of Triangles

De Moivre's Theorem

Main Page