Trigonometric Identities

Start from \tan\theta=\dfrac{\sin\theta}{\cos\theta} and \sin^2\theta+\cos^2\theta=1, and look for the statement that is a rearrangement of one of them.

1. \tan\theta=\dfrac{\sin\theta}{\cos\theta}; multiply both sides by \cos\theta: \cos\theta\tan\theta=\sin\theta
2. \cos\theta=1-\sin\theta is wrong: the identity is \cos^2\theta=1-\sin^2\theta, and you cannot square root term by term
3. \sin^2\theta-\cos^2\theta=1 is wrong: the identity has a plus sign
4. \tan\theta=\dfrac{\cos\theta}{\sin\theta} is upside down
5. Answer: \sin\theta=\cos\theta\tan\theta

Replace \tan\theta with \dfrac{\sin\theta}{\cos\theta}, then divide by a fraction by multiplying by its reciprocal.

1. \tan\theta=\dfrac{\sin\theta}{\cos\theta}
2. \sin\theta\div\dfrac{\sin\theta}{\cos\theta}=\sin\theta\times\dfrac{\cos\theta}{\sin\theta}
3. The \sin\theta cancels
4. Answer: \cos\theta

Take out the common factor 7 first.

1. 7\sin^2 52^\circ+7\cos^2 52^\circ=7(\sin^2 52^\circ+\cos^2 52^\circ)
2. \sin^2\theta+\cos^2\theta=1 for every angle, including 52^\circ
3. Answer: 7

Divide both sides by \cos\theta.

1. Divide both sides by \cos\theta: \dfrac{5\sin\theta}{\cos\theta}=2
2. \dfrac{\sin\theta}{\cos\theta}=\tan\theta, so 5\tan\theta=2
3. Answer: \tan\theta=\dfrac25

Take out the common factor, then look for \sin^2\theta+\cos^2\theta.

1. Common factor \cos\theta: \cos\theta(\cos^2\theta+\sin^2\theta)
2. \cos^2\theta+\sin^2\theta=1
3. So the expression is \cos\theta\times1
4. Answer: \cos\theta

Replace 1-\sin^2\theta using \sin^2\theta+\cos^2\theta=1, and write \tan\theta as \dfrac{\sin\theta}{\cos\theta}

1. 1-\sin^2\theta=\cos^2\theta
2. So the expression is \dfrac{\tan\theta\cos^2\theta}{\cos\theta}=\tan\theta\cos\theta
3. \tan\theta=\dfrac{\sin\theta}{\cos\theta}, so \tan\theta\cos\theta=\sin\theta
4. Answer: \sin\theta

Replace \cos^2\theta with 1-\sin^2\theta, in a bracket.

1. \cos^2\theta=1-\sin^2\theta
2. 5\sin^2\theta+2(1-\sin^2\theta)=5\sin^2\theta+2-2\sin^2\theta
3. Collect the \sin^2\theta terms
4. Answer: 2+3\sin^2\theta

Factorise the numerator and use \sin^2\theta+\cos^2\theta=1.

1. 3-3\cos^2\theta=3(1-\cos^2\theta)=3\sin^2\theta
2. So the fraction is \dfrac{3\sin^2\theta}{4\cos^2\theta}
3. \dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta
4. Answer: \dfrac34\tan^2\theta

Use \sin^2A=1-\cos^2A, then square root and simplify the surd.

1. \cos^2A=\dfrac{4}{49}
2. \sin^2A=1-\dfrac{4}{49}=\dfrac{45}{49}
3. A is acute, so \sin A is positive: \sin A=\dfrac{\sqrt{45}}{7}
4. \sqrt{45}=\sqrt9\times\sqrt5=3\sqrt5
5. Answer: \dfrac{3\sqrt5}{7}

Use \sin^2\theta+\cos^2\theta=1, then think about whether cosine is positive or negative for an obtuse angle.

1. \sin^2\theta=\dfrac{7}{16}
2. \cos^2\theta=1-\dfrac{7}{16}=\dfrac{9}{16}
3. \cos\theta=\pm\dfrac34
4. \theta is between 90^\circ and 180^\circ, where cosine is negative
5. Answer: \cos\theta=-\dfrac34

Replace \cos^2x with 1-\sin^2x, then collect every term on one side.

1. \cos^2x=1-\sin^2x, so 2(1-\sin^2x)=3\sin x
2. 2-2\sin^2x=3\sin x
3. Add 2\sin^2x to both sides and subtract 2: 0=2\sin^2x+3\sin x-2
4. Answer: 2\sin^2x+3\sin x-2=0

Check each line on its own against \sin^2\theta+\cos^2\theta=1 and \tan\theta=\dfrac{\sin\theta}{\cos\theta}.

1. Line 1 is right: 3 is a common factor of the numerator
2. Line 2 is right: 1-\sin^2\theta=\cos^2\theta, and \dfrac{\cos^2\theta}{\cos\theta}=\cos\theta
3. Line 3 is wrong: dividing by \dfrac{\sin\theta}{\cos\theta} means multiplying by \dfrac{\cos\theta}{\sin\theta}
4. Correctly, \dfrac{\sin\theta}{\tan\theta}=\cos\theta, and the left-hand side is 3\cos\theta+\cos\theta=4\cos\theta
5. Line 4 is wrong too, but only because of the error in line 3
6. Answer: Line 3

Find \sin\theta first, and decide whether it is positive or negative for an angle between 270^\circ and 360^\circ.

1. \sin^2\theta=1-\dfrac{64}{289}=\dfrac{225}{289}
2. \sin\theta=\pm\dfrac{15}{17}
3. Between 270^\circ and 360^\circ sine is negative, so \sin\theta=-\dfrac{15}{17}
4. \tan\theta=\dfrac{\sin\theta}{\cos\theta}=-\dfrac{15}{17}\div\dfrac{8}{17}
5. Answer: \tan\theta=-\dfrac{15}{8}

Replace \tan\theta, then write both terms over the common denominator \sin\theta.

1. \dfrac{\cos\theta}{\tan\theta}=\cos\theta\times\dfrac{\cos\theta}{\sin\theta}=\dfrac{\cos^2\theta}{\sin\theta}
2. \sin\theta=\dfrac{\sin^2\theta}{\sin\theta}
3. Add: \dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta}
4. \cos^2\theta+\sin^2\theta=1
5. Answer: \dfrac{1}{\sin\theta}

Factorise the numerator and the denominator: they share a bracket.

1. Numerator: \sin\theta(\cos\theta+\sin\theta)
2. Denominator: \cos\theta(\cos\theta+\sin\theta)
3. Cancel (\cos\theta+\sin\theta): \dfrac{\sin\theta}{\cos\theta}
4. Answer: \tan\theta

Factorise the first numerator and use \sin^2\theta+\cos^2\theta=1 on both numerators before dividing.

1. First numerator: \sin\theta(\sin^2\theta+\cos^2\theta)=\sin\theta
2. First fraction: \dfrac{\sin\theta}{\tan\theta}=\sin\theta\times\dfrac{\cos\theta}{\sin\theta}=\cos\theta
3. Second numerator: 2\cos^2\theta-2=-2(1-\cos^2\theta)=-2\sin^2\theta
4. Second fraction: \dfrac{-2\sin^2\theta}{\sin\theta}=-2\sin\theta
5. Answer: \cos\theta-2\sin\theta

Square both sides of the equation and use \sin^2\theta+\cos^2\theta=1. Answer part (a) before attempting part (b).

Part (a)

1. Square both sides: (\sin\theta-\cos\theta)^2=\dfrac19
2. Expand: \sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=\dfrac19
3. \sin^2\theta+\cos^2\theta=1, so 1-2\sin\theta\cos\theta=\dfrac19
4. 2\sin\theta\cos\theta=\dfrac89
5. Answer: \sin\theta\cos\theta=\dfrac49

Part (b)

1. Common denominator: \dfrac{1}{\sin\theta}-\dfrac{1}{\cos\theta}=\dfrac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}
2. \cos\theta-\sin\theta=-\dfrac13, and \sin\theta\cos\theta=\dfrac49 from part (a)
3. -\dfrac13\div\dfrac49=-\dfrac13\times\dfrac94
4. Answer: -\dfrac34

Write \tan x\cos x more simply and replace \cos^2x using \sin^2x+\cos^2x=1; the quadratic does not factorise. Answer part (a) before attempting part (b).

Part (a)

1. \tan x\cos x=\dfrac{\sin x}{\cos x}\times\cos x=\sin x
2. \cos^2x=1-\sin^2x
3. So 5(1-\sin^2x)+8\sin x=7, which is 5-5\sin^2x+8\sin x=7
4. Collect the terms on one side: 0=5\sin^2x-8\sin x+2
5. Answer: 5\sin^2x-8\sin x+2=0

Part (b)

1. Quadratic formula: \sin x=\dfrac{8\pm\sqrt{64-40}}{10}=\dfrac{8\pm\sqrt{24}}{10}
2. \sin x=1.2898\ldots or \sin x=0.3101\ldots
3. \sin x cannot be more than 1, so \sin x=0.3101\ldots
4. x=\sin^{-1}(0.3101\ldots)=18.06\ldots^\circ
5. Sine is symmetrical about 90^\circ: x=180^\circ-18.06\ldots^\circ=161.93\ldots^\circ
6. Answer: x=18.1^\circ,\ 161.9^\circ

Divide the equation by \cos\theta, and remember which ratios are positive for an obtuse angle. Answer part (a) before attempting part (b).

Part (a)

1. 3\sin\theta=-2\cos\theta
2. Divide both sides by 3\cos\theta: \dfrac{\sin\theta}{\cos\theta}=-\dfrac23
3. Answer: \tan\theta=-\dfrac23

Part (b)

1. From the equation, \cos\theta=-\dfrac32\sin\theta
2. Substitute into \sin^2\theta+\cos^2\theta=1: \sin^2\theta+\dfrac94\sin^2\theta=1
3. \dfrac{13}{4}\sin^2\theta=1, so \sin^2\theta=\dfrac{4}{13}
4. \theta is obtuse, so \sin\theta is positive: \sin\theta=\dfrac{2}{\sqrt{13}}
5. Answer: \sin\theta=\dfrac{2}{\sqrt{13}}=\dfrac{2\sqrt{13}}{13}

Part (c)

1. \tan^{-1}\left(\dfrac23\right)=33.69\ldots^\circ
2. \tan\theta is negative and \theta is obtuse, so \theta=180^\circ-33.69\ldots^\circ
3. =146.30\ldots^\circ
4. Answer: \theta=146.3^\circ

Write the denominator as \cos^2\theta and split the 5 in the numerator as 5(\sin^2\theta+\cos^2\theta); when you square root, remember the negative value too.

Part (a)

1. Denominator: 1-\sin^2\theta=\cos^2\theta
2. Numerator: 8\sin^2\theta-5=8\sin^2\theta-5(\sin^2\theta+\cos^2\theta)=3\sin^2\theta-5\cos^2\theta
3. Divide each term by \cos^2\theta: \dfrac{3\sin^2\theta}{\cos^2\theta}-\dfrac{5\cos^2\theta}{\cos^2\theta}
4. \dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta
5. Answer: 3\tan^2\theta-5

Part (b)

1. Using part (a): 3\tan^2\theta-5=7
2. 3\tan^2\theta=12, so \tan^2\theta=4
3. \tan\theta=2 or \tan\theta=-2
4. \tan\theta=2: \theta=63.43\ldots^\circ and 180^\circ+63.43\ldots^\circ=243.43\ldots^\circ
5. \tan\theta=-2: \theta=180^\circ-63.43\ldots^\circ=116.56\ldots^\circ and 360^\circ-63.43\ldots^\circ=296.56\ldots^\circ
6. Answer: 63.4^\circ,\ 116.6^\circ,\ 243.4^\circ,\ 296.6^\circ