Question

1) Use the compound-angle formulae to write the following as surds.

(i) sin 75° = sin(45° +30°)

(ii) cos 135º = cos(90° +45°)

Answer 1

Problem 1:

`sin(75^o)=sin(45^o+30^o)\\sin(A+B)=sinAcosB+cosAsinB\\sin(45^o+30^o)=sin45^ocos30^o+cos45^osin30^o\\(1)/(√(2) ) *(√(3) )/(2) +(1)/(√(2) ) ]*(1)/(2) \\(√(3) )/(2√(2) ) +(1)/(2√(2) ) ---- > (√(3)+1)/(2√(2) ) (ANS)`

Problem 2:

`cos135^o=-cos45^o=-(1)/(√(2) )`

Step-by-step explanation:

Cos 135° is an angle in the second quadrant.

In the second quadrant, cos is negative.
`cos\theta=(x)/(r)`

`cos135=cos(180-45)=-cos45^o`

An angle of 45° is found in a right-angled triangle of sides
`1:1:√(2)`

`cos45^o=(1)/(√(2) )`

`cos135^o=-cos45^o=-(1)/(√(2) )`

Note that
`√(2)` is an irrational number and cannot be given as an exact decimal.

Little messy math, but thanks for asking the question.

- Eddie