Exact values, any angle and inverse trig functions

What you need to be able to do

What you should already know

Exact values from two triangles

Beyond 90°: sin and cos on a circle

A right-angled triangle can only hold angles below 90°, so for larger angles we need a better definition. Put a point \(P\) on a circle of radius 1, centred at the origin, and turn the radius \(OP\) anticlockwise from the positive \(x\)-axis through the angle \(\theta\). Then

For acute angles this is SOH CAH TOA with hypotenuse 1. But it works for every angle, and it explains everything that follows: at \(90°\), \(P=(0,1)\) so \(\cos90°=0\), \(\sin90°=1\) and \(\tan90°\) is undefined (dividing by 0); at \(180°\), \(P=(-1,0)\); in the 2nd quadrant \(x<0\) and \(y>0\), so cos is negative and sin positive. The signs in each quadrant are just the signs of the coordinates.

Any angle: the related angle and CAST

Inverse functions and principal values

\(\sin^{-1}x\) means "the angle whose sine is \(x\)". Many angles have the same sine, so the calculator gives just one, the principal value, from a fixed range:

1. \(\cos^{-1}\!\left(\tfrac{\sqrt3}2\right)=\tfrac\pi6\), so \(-\cos^{-1}\!\left(\tfrac{\sqrt3}2\right)=-\tfrac\pi6\) (the minus is outside).
2. \(\tan^{-1}(\text{something})=-\tfrac\pi4\) means \(\text{something}=\tan\!\left(-\tfrac\pi4\right)=-1\): take tan of both sides.
3. \(\cos^{-1}(-\tfrac12)=\tfrac{2\pi}3\) (2nd quadrant, inside \(0\) to \(\pi\)); \(\sin^{-1}(-\tfrac12)=-\tfrac\pi6\) (not \(\tfrac{7\pi}6\): outside the range).

Given one ratio, find the others

1. Draw the right-angled triangle for the acute related angle: e.g. \(\cos\alpha=\tfrac5{13}\) gives adjacent 5, hypotenuse 13, opposite \(\sqrt{13^2-5^2}=12\).
2. Or use the identity \(\sin^2\theta+\cos^2\theta=1\): \(\sin\alpha=\sqrt{1-\tfrac{25}{169}}=\tfrac{12}{13}\). Either way is fine; show one of them.
3. Fix the sign from the quadrant: if \(90°<\beta<180°\) and \(\sin\beta=a\), then \(\cos\beta=-\sqrt{1-a^2}\) (cos is negative in the 2nd quadrant) and \(\tan\beta=\dfrac{a}{-\sqrt{1-a^2}}\).
4. Letters work the same way: \(\tan^2\beta=\dfrac{\sin^2\beta}{\cos^2\beta}=\dfrac{a^2}{1-a^2}\).

Case by case

Key facts at a glance

• Exact values come from two triangles: 1, 1, \(\sqrt2\) and 1, \(\sqrt3\), 2.
• On the unit circle \(\cos\theta=x\), \(\sin\theta=y\): this defines them for every angle.
• Any angle: related acute angle for the size, CAST for the sign.
• Principal values: \(\sin^{-1}\) in \([-\tfrac\pi2,\tfrac\pi2]\), \(\cos^{-1}\) in \([0,\pi]\), \(\tan^{-1}\) in \((-\tfrac\pi2,\tfrac\pi2)\).
• Given one ratio: a right-angled triangle or \(\sin^2+\cos^2=1\), then the sign from the quadrant.

How this is examined

"Exact" and "without using the trigonometric functions on your calculator" mean the working must be seen: "Correct answer with no working shown scores 0". Schemes give the method mark for a decimal but the A1 only for the exact form, and condone a degree-mode value for the method mark only.

Mark-scheme focus

Show, without using a calculator, that …

Credited

• multiply through by \(\cos x\): \(4\sin x+3\cos^2x+1=0\) M1
• \(\cos^2x=1-\sin^2x\) and simplify to a 3-term quadratic: \(3\sin^2x-4\sin x-4=0\) M1
• \((3\sin x+2)(\sin x-2)=0\Rightarrow\sin x=-\tfrac23\) (\(\sin x=2\) impossible) {A1 AG}

Not credited

• finding \(x\) on a calculator and checking it
• \(\sin x=-\tfrac23\) stated after the quadratic with no factorising or reason to reject 2
• \(\tan x\) left in the equation (it must become \(\tfrac{\sin x}{\cos x}\))

Why

"Show" and "without a calculator" mean every algebraic step is marked. The answer is given (AG), so the marks are all for the route; rejecting sin x = 2 completes it.

From: 9709/11/O/N/19 Q5(i).

Worked examples

Example 1 · Exact value without a calculator (Paper 1, 5 marks) ★ Easy

1 Read it with me what is actually happening?

\(\alpha\) is an angle whose cosine is \(\tfrac8{17}\). It lies between 0 and 90° (principal value, positive cos). We need its sin and tan as fractions, then add two fractions.

• cos⁻¹(8/17)an acute angle with cos α = 8/17
• without using the trigonometric functionsevery value must come from a triangle or an identity
• exacta fraction, not a decimal

2 What is it asking for? the target, part by part

• \(\tfrac1{\sin\alpha}+\tfrac1{\tan\alpha}\) exactly5 marks

3 What do we know? everything the question gives us

• \(\cos\alpha=\tfrac8{17}\)adjacent 8, hypotenuse 17

• 8, 15, 17 is a Pythagorean triple: \(8^2+15^2=17^2\).

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Find\(\tfrac1{\sin\alpha}+\tfrac1{\tan\alpha}\)
  • sin αtriangle or sin² + cos² = 1
  • tan αopposite ÷ adjacent

What is the missing side?

\(\sqrt{289-64}=15\).

Is sin α positive?

Yes: \(\alpha\) is acute (cos⁻¹ of a positive number).

5 Now solve it have a go on paper first, then open the worked solution

1. \(\sin\alpha=\sqrt{1-\tfrac{64}{289}}=\tfrac{15}{17}\) *M1 A1
2. \(\tan\alpha=\tfrac{15/17}{8/17}=\tfrac{15}8\) A1
3. \(\tfrac1{\sin\alpha}+\tfrac1{\tan\alpha}=\tfrac{17}{15}+\tfrac8{15}=\tfrac{25}{15}\) DM1
4. \(=\tfrac53\) A1

6 Check it does the answer make sense?

Calculator check (allowed now that we're done): \(\alpha=61.9°\), \(1/\sin\alpha+1/\tan\alpha=1.133+0.533=1.667\) ✓.

Example 2 · A ratio given as a letter, angle obtuse (Paper 1, 3 marks) ★★ Medium

1 Read it with me what is actually happening?

\(\beta\) is obtuse, so it's in the 2nd quadrant: sin positive, cos negative. Write everything in terms of \(\sin\beta=a\) and \(\cos\beta\), then replace \(\cos\beta\) using the identity, with the right sign.

• between 90° and 180°2nd quadrant: cos β < 0
• in terms of athe answer contains only a

2 What is it asking for? the target, part by part

• (a)\(\tan^2\beta-3\sin\beta\cos\beta\)3 marks

3 What do we know? everything the question gives us

• \(\sin\beta=a\)

• \(\cos\beta=-\sqrt{1-a^2}\): the minus sign is a mark on its own.

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findthe expression in a
  • Use\(\tan^2\beta=\tfrac{\sin^2\beta}{\cos^2\beta}\)
  • cos β\(-\sqrt{1-a^2}\)
  • Trap\(+\sqrt{1-a^2}\)

What is \(\cos^2\beta\)?

\(1-a^2\) (the sign doesn't matter once it is squared).

What is \(-3\sin\beta\cos\beta\)?

\(-3a\cdot(-\sqrt{1-a^2})=+3a\sqrt{1-a^2}\).

5 Now solve it have a go on paper first, then open the worked solution

1. \(\tan^2\beta=\dfrac{\sin^2\beta}{\cos^2\beta}=\dfrac{a^2}{1-a^2}\) B1
2. \(\cos\beta=-\sqrt{1-a^2}\) B1
3. \(\tan^2\beta-3\sin\beta\cos\beta=\dfrac{a^2}{1-a^2}+3a\sqrt{1-a^2}\) B1

6 Check it does the answer make sense?

Try \(\beta=150°\), \(a=\tfrac12\): \(\tan^2150°=\tfrac13\) and \(-3\sin150°\cos150°=\tfrac{3\sqrt3}4\). Formula: \(\tfrac{1/4}{3/4}+\tfrac32\cdot\tfrac{\sqrt3}2=\tfrac13+\tfrac{3\sqrt3}4\) ✓.

Example 3 · Coordinates on a cosine curve, then an inverse equation (Paper 1, 5 marks) ★★ Medium

1 Read it with me what is actually happening?

The curve is \(y=k\cos x\) moved \(\tfrac\pi6\) to the right. Count along the picture: max, zero, min, zero (A), max, zero, min (B). Zeros of cos are at \(\tfrac\pi2,\tfrac{3\pi}2,\ldots\); minima at \(\pi,3\pi,\ldots\). Then add \(\tfrac\pi6\).

• crosses the x-axis at Ay = 0: the 2nd zero after the start
• B is a minimum pointy = −k, the 2nd minimum
• exactmultiples of π

2 What is it asking for? the target, part by part

• (a)A and B3 marks
• (b)exact t2 marks

3 What do we know? everything the question gives us

• \(y=k\cos(x-\tfrac\pi6)\)k > 0

• A is the zero where the curve goes up: after the first minimum.

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)A, B
  • Zeros\(x-\tfrac\pi6=\tfrac{3\pi}2\Rightarrow x=\tfrac{5\pi}3\)
  • 2nd minimum\(x-\tfrac\pi6=3\pi\Rightarrow x=\tfrac{19\pi}6\)
• (b)t
  • Known\(\cos^{-1}\tfrac{\sqrt2}2=\tfrac\pi4\)

Which zero is A?

The curve starts near a max, falls through \(x=\tfrac{2\pi}3\), reaches a min at \(\tfrac{7\pi}6\) and rises through A.

What is the y-coordinate of B?

\(-k\): the least value of \(k\cos(\ldots)\).

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(A=\left(\tfrac53\pi,\ 0\right)\) B1
2. \(B\): \(x=\tfrac{19}6\pi\) B1, \(y=-k\) B1
3. (b) \(3\sin^{-1}(3t)=\pi-2\cdot\tfrac\pi4=\tfrac\pi2\), \(\sin^{-1}(3t)=\tfrac\pi6\) M1
4. \(3t=\tfrac12\Rightarrow t=\tfrac16\) A1

6 Check it does the answer make sense?

\(\tfrac{19\pi}6-\tfrac{5\pi}3=\tfrac{9\pi}6=\tfrac{3\pi}2\): from a rising zero to the next-but-one minimum is ¾ of a period (2π) ✓.

Summary

• Two triangles (1, 1, √2 and 1, √3, 2) give every exact value of 30°, 45°, 60°.
• Any angle: related acute angle + sign from CAST (All, Sin, Tan, Cos).
• Principal values: \(\sin^{-1}\) and \(\tan^{-1}\) give \(-\tfrac\pi2\) to \(\tfrac\pi2\); \(\cos^{-1}\) gives \(0\) to \(\pi\).
• Given one ratio: triangle or \(\sin^2+\cos^2=1\), then the sign from the quadrant.

Can I do this?

Quiz

Concept checks

Answer each one in your head first, then open it.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.