# trigonometric identities problems

(2+1)(2−1)cos⁡θ=(2+1)sin⁡θ(2−1)cos⁡θ=2sin⁡θ+sin⁡θ⇒cos⁡θ−sin⁡θ=2sin⁡θ. Knowing that cos α = ¼ , and that 270º < α < 360°, calculate the remaining trigonometric ratios of angle α. \sin \theta &= \cos \left( \frac{\pi}{2}-\theta \right) \\ \cot \dfrac{\pi}{16} \cdot \cot \dfrac{2 \pi}{16} \cdot \cot \dfrac{3 \pi}{16} \times \cdots \times \cot \dfrac{7 \pi}{16} &=\underbrace{\sin^2 \dfrac{\pi}{10} + \cos^2 \dfrac{\pi}{10}}_{1} + \underbrace{\cos^2 \dfrac{\pi}{10} + \sin^2 \dfrac{\pi}{10}}_{1} \\ A  =  (1 - cos Î¸)(1 + cos Î¸)(1 + cot2Î¸), A  =  sin2Î¸  + sin2Î¸ â (cos2Î¸/sin2Î¸). (tanθ+cotθ)2​=tan2θ+cot2θ+2tanθcotθ=tan2θ+cot2θ+2=(1+tan2θ)+(1+cot2θ)=sec2θ+csc2θ​, 2. sec⁡2θ+csc⁡2θ=1cos⁡2θ+1sin⁡2θ=sin⁡2θ+cos⁡2θsin⁡2θ⋅cos⁡2θ=1sin⁡2θ⋅cos⁡2θ=sec⁡2θ⋅csc⁡2θ. \hline /2, calculate the remaining trigonometric ratios of angle α. Donate or volunteer today! \sin^2 \dfrac{\pi}{10} + \sin^2 \dfrac{4\pi}{10} + \sin^2 \dfrac{6 \pi}{10} + \sin^2 \dfrac{9 \pi}{10} sin⁡2A+cos⁡2A=1tan⁡2A+1=sec⁡2Acot⁡2A+1=csc⁡2A.\begin{aligned} \sin^2 A + \cos^2 A &=& 1 \\ \tan^2 A + 1 &=& \sec^2 A \\ \cot^2 A + 1 &=& \csc^2 A. \hline A  =  (cos2Î¸/sin Î¸ cos Î¸) + (sin2Î¸/sin Î¸ cos Î¸), cos Î¸/(1 - tan Î¸) + sin Î¸/(1 - cot Î¸)  =  sin Î¸ + cos Î¸, Let A  =  cos Î¸/(1 - tan Î¸) + sin Î¸/(1 - cot Î¸)  and, A  =  cos Î¸/{1 - (sin Î¸/cos Î¸)} + sin Î¸/{1 - (cos Î¸/sin Î¸)}, A  =  cos2Î¸/(cos Î¸ - sin Î¸) + sin2Î¸/(sin Î¸ - cos Î¸), A  =  cos2Î¸/(cos Î¸ - sin Î¸) - sin2Î¸/(cos Î¸ - sin Î¸), A  =  (cos2Î¸ - sin2Î¸) / (cos Î¸ - sin Î¸), A  =  [(cos Î¸ + sin Î¸)(cos Î¸ - sin Î¸)] / (cos Î¸ - sin Î¸). 52=(3sin⁡θ+4cos⁡θ)2a2=(4sin⁡θ−3cos⁡θ)252+a2=(3sin⁡θ+4cos⁡θ)2+(4sin⁡θ−3cos⁡θ)2a2+25=9sin⁡2θ+16cos⁡2θ+24sin⁡θcos⁡θ+16sin⁡2θ+9cos⁡2θ−24sin⁡θcos⁡θa2+25=25(sin⁡2θ+cos⁡2θ)a2+25=25a=0⇒4sin⁡θ−3cos⁡θ=0. New user? & = 2\big(1 - 3\sin^2 \theta \cos^2 \theta\big) - 3\big(1 - 2\sin^2 \theta \cos^2 \theta\big) \\ □​​. \end{aligned}cosθsinθcotθcscθ​=sin(2π​−θ)=cos(2π​−θ)=tan(2π​−θ)=sec(2π​−θ).​, sin⁡2θ=2sin⁡θcos⁡θcos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θtan⁡2θ=2tan⁡θ1−tan⁡2θ.\begin{aligned} \sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{-5}{13}.sinθ=secθtanθ​=13−5​. & = 2\left[\big(\sin^2 \theta\big)^3 + \big(\cos^2 \theta\big)^3\right] - 3\left[\big(\sin^2 \theta\big)^2 + \big(\cos^2 \theta\big)^2\right] \\ If you're seeing this message, it means we're having trouble loading external resources on our website. \hline

\end{aligned}1. \tan(x-y) &= \dfrac{\tan x - \tan y}{1 + \tan x \tan y}. Log in. Let A  =  (1 - cos Î¸)(1 + cos Î¸)(1 + cot2Î¸)  =  1 and B  =  1. \sin \theta &=\sin(\theta+2\pi) &\quad \csc \theta &=\csc(\theta+2\pi)\\ \end{array}θsinθcosθtanθ​0∘20​​24​​0​6π​=30∘21​​23​​3​1​​4π​=45∘22​​22​​1​3π​=60∘23​​21​​3​​2π​=90∘24​​20​​∞​​, cos⁡2θ+sin⁡2θ=11+tan⁡2θ=sec⁡2θcot⁡2θ+1=csc⁡2θ.\begin{aligned} â{(sec Î¸ â 1)/(sec Î¸ + 1)}  =  cosec Î¸ - cot Î¸. Students are taught about trigonometric identities in school and are an important part of higher-level mathematics. Identities In Use‎ > ‎ Word Problem During a stormy night in Louisiana, a tree fell on a residents home. Domain and range of trigonometric … Let A  =  (1 - sin A)/(1 + sin A) and B  =  (sec A - tan A)2. \tan^2 (x+y)?tan2(x+y)? Trigonometric Identities. \end{aligned} sin2θcos2θ​=21​(1−cos2θ)=21​(1+cos2θ).​, cos⁡xcos⁡y=12(cos⁡(x−y)+cos⁡(x+y))sin⁡xcos⁡y=12(sin⁡(x−y)+sin⁡(x+y))cos⁡xsin⁡y=12(sin⁡(x+y)−sin⁡(x−y))sin⁡xsin⁡y=12(cos⁡(x−y)−cos⁡(x+y)).\begin{aligned} □\begin{aligned}

Proving Trigonometric Identities Calculator Get detailed solutions to your math problems with our Proving Trigonometric Identities step-by-step calculator. Knowing that tan α = 2, and that 180º < α <270°, calculate the remaining trigonometric ratios of angle α. Trigonometric ratios of supplementary angles Trigonometric identities Problems on trigonometric identities Trigonometry heights and distances &= 1.\ _\square When working with trigonometric identities, it may be useful to keep the following tips in mind: The fundamental period of the graphs of sin⁡x,cos⁡x,csc⁡x,sec⁡x\sin x, \cos x, \csc x, \sec x sinx,cosx,cscx,secx is 2π,2\pi,2π, while the fundamental period of the graphs of tan⁡x,cot⁡x \tan x, \cot x tanx,cotx is π\piπ.

Our mission is to provide a free, world-class education to anyone, anywhere. \end{aligned}cot16π​⋅cot162π​⋅cot163π​×⋯×cot167π​​=(cot16π​⋅cot167π​)⋅(cot162π​⋅cot166π​)⋅(cot163π​⋅cot165π​)⋅cot164π​=(cot16π​⋅cot(2π​−16π​))⋅(cot162π​⋅cot(2π​−162π​))⋅(cot163π​⋅cot(2π​−163π​))⋅cot4π​=1(cot16π​⋅tan16π​)​​⋅1(cot162π​⋅tan162π​)​​⋅1(cot163π​⋅tan163π​)​​⋅1=1.

Let A  =  sec Î¸ â(1 - sin2Î¸)  and B  =  1. \cos x \cos y & = \frac{1}{2} \big(\cos (x - y) + \cos(x+ y) \big) \\ From this we get to know that sec⁡θ\sec \thetasecθ is positive and tan⁡θ\tan \thetatanθ is negative. \Rightarrow 4\sin\theta - 3\cos\theta & = 0.\ _\square

Trigonometric Identities. Identities In Use‎ > ‎ Word Problem During a stormy night in Louisiana, a tree fell on a residents home. a^2 + 25 & = 25(\sin^2 \theta + \cos^2 \theta) \\ & = \sec^2 \theta \cdot \csc^2 \theta.\ _\square \cot(-\theta) &= -\cot \theta\\ \tan \theta & 0 & \frac { 1}{\sqrt{3} } & 1 & \sqrt{3} & \infty \\ Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. \cot^2 \theta + 1 &= \csc^2 \theta. \end{aligned}(2​+1)(2​−1)cosθ(2−1)cosθ⇒cosθ−sinθ​=(2​+1)sinθ=2​sinθ+sinθ=2​sinθ. \end{aligned}cos(−θ)sin(−θ)tan(−θ)cot(−θ)csc(−θ)sec(−θ)​=cosθ=−sinθ=−tanθ=−cotθ=−cscθ=secθ.​, sin⁡θ=sin⁡(θ+2π)csc⁡θ=csc⁡(θ+2π)cos⁡θ=cos⁡(θ+2π)sec⁡θ=sec⁡(θ+2π)tan⁡θ=tan⁡(θ+π)cot⁡θ=cot⁡(θ+π).\begin{aligned} Calculate the trigonometric ratios of 15 (from the 45º and 30º). 1 + \tan^2 \theta &= \sec^2 \theta \\ I like to spend my time reading, gardening, running, learning languages and exploring new places. Practice your math skills and learn step by step with our math solver. Find the value of 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)2\big(\sin^6 \theta + \cos^6 \theta\big) - 3\big(\sin^4 \theta + \cos^4 \theta\big)2(sin6θ+cos6θ)−3(sin4θ+cos4θ).

Knowing that tan α = 2, and that 180º < α < 270°, calculate the remaining trigonometric ratios of angle α. When working with trigonometric identities, it may be useful to keep the following tips in mind: Draw a picture illustrating the problem if it involves only the basic trigonometric functions.

If 3sin⁡θ+4cos⁡θ=53 \sin \theta + 4 \cos \theta = 53sinθ+4cosθ=5, then find the value of 4sin⁡θ−3cos⁡θ4 \sin \theta - 3 \cos \theta4sinθ−3cosθ. sin⁡2π10+sin⁡24π10+sin⁡26π10+sin⁡29π10=sin⁡2(π10)+sin⁡2(π2−π10)+sin⁡2(π2+π10)+sin⁡2(π−π10)=sin⁡2π10+cos⁡2π10⏟1+cos⁡2π10+sin⁡2π10⏟1=1+1=2. \sin 3 \theta &= 3 \sin \theta - 4 \sin ^3 \theta \\

\end{aligned} sin2θcos2θtan2θ​=2sinθcosθ=cos2θ−sin2θ=2cos2θ−1=1−2sin2θ=1−tan2θ2tanθ​.​, sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡ysin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡ycos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡ycos⁡(x−y)=cos⁡xcos⁡y+sin⁡xsin⁡ytan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡ytan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y.\begin{aligned} Sitemap. \sin x+\sin y & =2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ & = \tan^2 \theta + \cot^2 \theta + 2 \\ \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ \end{aligned} sin2A+cos2Atan2A+1cot2A+1​===​1sec2Acsc2A.​, sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B.\begin{aligned} \sin(A\pm B) &=& \sin A \cos B \pm \cos A \sin B \\ \cos(A\pm B) &=& \cos A \cos B \mp \sin A \sin B \\ \tan(A\pm B) &=& \frac{ \tan A \pm \tan B } { 1 \mp \tan A \tan B }. \sin x \sin y &= \frac{1}{2} \big(\cos (x - y) - \cos(x + y) \big). &=1 + 1 \\ sin⁡2θ+cos⁡2θ=1cos⁡2θ=1−sin⁡2θ=1−1625=925⇒cos⁡θ=±35⇒cos⁡θ=−35. \sin \theta & \frac {\sqrt{0}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{4}} {2} \\ Given that cos⁡35°=α\cos{35°}=\alphacos35°=α, express sin⁡2015∘\sin{2015^\circ}sin2015∘ in terms of α\alphaα.

Sitemap. Let A  =  cot Î¸ + tan Î¸ and B  =  sec Î¸ csc Î¸. \theta & 0^\circ & \frac{\pi}{6} = 30^\circ & \frac{\pi}{4} = 45^\circ & \frac{\pi}{3} = 60^\circ & \frac{\pi}{2} = 90^\circ\\

Add rational expressions by finding common denominators. Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse  trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Hcf and Lcm of Fractions - Formulas - Examples, PROBLEMS ON TRIGONOMETRIC IDENTITIES WITH SOLUTIONS, =  â[{(sec Î¸ - 1) (sec Î¸ - 1)}/{(sec Î¸ + 1) (sec Î¸ - 1)}], (1 - sin A)/(1 + sin A)  =  (sec A - tan A). \end{aligned}cosxcosysinxcosycosxsinysinxsiny​=21​(cos(x−y)+cos(x+y))=21​(sin(x−y)+sin(x+y))=21​(sin(x+y)−sin(x−y))=21​(cos(x−y)−cos(x+y)).​, sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)cos⁡x+cos⁡y=2cos⁡(x+y2)cos⁡(x−y2).\begin{aligned}

\hline Trigonometric ratios of angles greater than or equal to 360 degree. \Rightarrow \cos\theta - \sin\theta & = \sqrt2 \sin\theta.\ _\square To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Learn how to solve trigonometric equations and how to use trigonometric identities to solve various problems. Let A  =  tan4Î¸ + tan2Î¸  and B  =  sec4Î¸ + sec2Î¸. sin⁡xsin⁡y=12,cos⁡xcos⁡y=32, \frac {\sin x}{\sin y } = \frac {1}{2}, \quad \frac {\cos x}{\cos y } = \frac 3 2 , sinysinx​=21​,cosycosx​=23​. \end{aligned}sinθcosθtanθ​=sin(θ+2π)=cos(θ+2π)=tan(θ+π)​cscθsecθcotθ​=csc(θ+2π)=sec(θ+2π)=cot(θ+π).​, cos⁡θ=sin⁡(π2−θ)sin⁡θ=cos⁡(π2−θ)cot⁡θ=tan⁡(π2−θ)csc⁡θ=sec⁡(π2−θ).
□​​. \cos \theta & = \sin \left( \frac{\pi}{2} - \theta \right) \\