Exact Trig Values - Higher

Recall the sine values for the three special acute angles.

1. The exact sine values at 30^\circ, 45^\circ and 60^\circ are \frac12, \frac{\sqrt2}{2} and \frac{\sqrt3}{2}.
2. Therefore \sin60^\circ=\frac{\sqrt3}{2}.

Draw a right-angled triangle with angles 30^\circ and 60^\circ, and sides 1, \sqrt3 and 2.

1. Use half of an equilateral triangle of side 2: it has angles 30^\circ, 60^\circ and 90^\circ, hypotenuse 2, shortest side 1 and height \sqrt{2^2-1^2}=\sqrt3.
2. The perpendicular splits the equilateral triangle into two congruent right triangles. In the blue half, \sqrt3 is adjacent to 30^\circ and 2 is the hypotenuse.

3. The side next to the 30^\circ angle is \sqrt3, so \cos30^\circ=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt3}{2}.
4. The option \frac12 is \cos60^\circ, not \cos30^\circ.

Tangent is the opposite side divided by the adjacent side in a right-angled triangle.

1. In the 30^\circ–60^\circ–90^\circ triangle with sides 1, \sqrt3 and 2, the side opposite 30^\circ is 1 and the side adjacent to it is \sqrt3.
2. From the marked 30^\circ angle, the opposite side is 1 and the adjacent side is \sqrt3.

3. \tan30^\circ=\frac{\text{opposite}}{\text{adjacent}}=\frac1{\sqrt3}.
4. Rationalise the denominator: \frac1{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{3}.
5. The option \sqrt3 is \tan60^\circ, the other way up.

Match the ratio to an exact sine value.

1. \sin45^\circ=\frac{\sqrt2}{2}.
2. As x is acute, x=45^\circ.

You need \sin30^\circ, \cos0^\circ and \tan45^\circ.

1. The exact values are \sin30^\circ=\frac12, \cos0^\circ=1 and \tan45^\circ=1.
2. Substitute: 5\times\frac12+2\times1-1.
3. =\frac52+2-1=\frac72 (or 3.5).

Both trigonometric values involve the same surd.

1. Both \cos30^\circ and \sin60^\circ equal \frac{\sqrt3}{2}.
2. \frac{3\sqrt3}{2}-\frac{\sqrt3}{2}=\sqrt3.

Square each exact value separately.

1. (\sin45^\circ)^2=(\frac{\sqrt2}{2})^2=\frac12.
2. (\cos30^\circ)^2=(\frac{\sqrt3}{2})^2=\frac34.
3. The sum is \frac12+\frac34=\frac54.

Dividing by one half doubles a value.

1. \tan60^\circ=\sqrt3 and \sin30^\circ=\frac12.
2. \sqrt3\div\frac12=2\sqrt3.

Use the exact sine of the given angle.

1. \sin30^\circ=\frac{BC}{12}.
2. BC=12\times\frac12=6 cm.

Choose the ratio linking the adjacent side and the hypotenuse.

1. \cos60^\circ=\frac7{AC}.
2. As \cos60^\circ=\frac12, AC=14 cm.

The perpendicular splits the equilateral triangle into two equal right-angled triangles.

1. The perpendicular from C cuts the equilateral triangle into two right-angled triangles, so AM=4 cm and angle CAM=60^\circ.
2. In right-angled triangle AMC, the hypotenuse is AC=8 cm and CM is opposite the 60^\circ angle, so CM=8\sin60^\circ.
3. CM=8\times\frac{\sqrt3}{2}=4\sqrt3 cm.

Use the exact tangent and cosine to find the two missing sides.

1. BC=9\tan45^\circ=9 cm.
2. AC=\frac9{\cos45^\circ}=9\sqrt2 cm.
3. The perimeter is 9+9+9\sqrt2=18+9\sqrt2 cm.

Use the area formula with the two sides enclosing the given angle.

1. \text{Area}=\frac12\times6\times10\sin60^\circ.
2. This is 30\times\frac{\sqrt3}{2}=15\sqrt3 cm^2.

Use sine, then simplify the product of the two surds.

1. BC=10\sqrt3\sin60^\circ.
2. BC=10\sqrt3\times\frac{\sqrt3}{2}=15 cm.

Add the height of the square to the perpendicular height of the equilateral triangle.

1. The triangular roof has height 6\sin60^\circ=3\sqrt3 cm.
2. The square has height 6 cm.
3. The total height is 6+3\sqrt3 cm.

Evaluate each product before subtracting.

1. 3\sin60^\circ\tan30^\circ=3\times\frac{\sqrt3}{2}\times\frac{\sqrt3}{3}=\frac32.
2. 2(\cos45^\circ)^2=2\times\frac12=1.
3. The result is \frac32-1=\frac12.

Use the cosine rule with the angle at A, which is opposite the 7 cm side.

Part (a)

1. Angle BAC is opposite BC, so use the cosine rule BC^2=AB^2+AC^2-2\times AB\times AC\cos A.
2. 7^2=8^2+5^2-2\times8\times5\cos A, so 49=89-80\cos A.
3. 80\cos A=40, so \cos A=\frac12.
4. The angle whose cosine is \frac12 is 60^\circ, so angle BAC=60^\circ.

Part (b)

1. Use Area =\frac12ab\sin C with the two sides either side of angle A: Area =\frac12\times8\times5\times\sin60^\circ.
2. \sin60^\circ=\frac{\sqrt3}{2}, so Area =20\times\frac{\sqrt3}{2}.
3. Area =10\sqrt3 cm^2.

Find the area of each right-angled triangle, using exact trigonometric values.

1. In triangle ABC, AB=8\cos30^\circ=4\sqrt3 cm and BC=8\sin30^\circ=4 cm.
2. Its area is \frac12\times4\sqrt3\times4=8\sqrt3 cm^2.
3. In triangle ACD, CD=8\tan45^\circ=8 cm, so its area is \frac12\times8\times8=32 cm^2.
4. The total area is 32+8\sqrt3 cm^2.

Use the sine rule with the known side opposite the known angle.

1. \frac{BC}{\sin30^\circ}=\frac{10}{\sin45^\circ}.
2. BC=\frac{10\times\frac12}{\frac{\sqrt2}{2}}=\frac{10}{\sqrt2}.
3. Rationalising gives BC=5\sqrt2 cm.

Find the base diagonal exactly, then recognise an exact tangent ratio.

1. AC^2=3^2+(3\sqrt3)^2=9+27=36, so AC=6 cm.
2. The projection of AG onto the base is AC, so the required angle is \angle GAC.
3. \tan(\angle GAC)=\frac{CG}{AC}=\frac{6\sqrt3}{6}=\sqrt3.
4. Therefore the angle is 60^\circ.