Show how '2sin(x)+sec(x+ π/6)=0' can be expressed as √3sin(x)cos(x)+cos^2(x)=0.

To do this question we must first be aware of a few trigonometric identities. 1. sec(x) = 1/cos(x). 2. cos(A + B) = cos(A)cos(B) - sin(A)sin(B). (double angle formula) 3 . sin^2(x) = 1 - cos^2(x) Before diving in, inspect the given expression and expression you are trying to get to. All terms on the RHS have 'x' within the brackets, therefore we must be using double angle formula to separate a sec term on the LHS. Only sin and cos are present on the RHS therefore we must try to change the sec term into another trig function. 1. Using identity 1. above, '2sin(x)+sec(x+ π/6)=0' can be changed to '2sin(x)+1/cos(x + π/6)=0' 2. considering only cos(x + π/6), use identity 2. to express it as functions only containing 1 term. A = x B = π/6 cos(x + π/6) = cos(x)cos(π/6) - sin(x)sin(π/6) here it is useful to notice that cos(π/6) and sin(π/6) are known values. Put each one into your calculator and you will find, cos(π/6) = √3/2 and sin(π/6) = 1/2 therefore 2sin(x)+sec(x+ π/6)=0 → 2sin(x)+1/((√3/2)cos(x) - (1/2)sin(x)) = 0 3. In order to get it into a form closer to the final expression, multiply through by (√3/2)cos(x) - (1/2)sin(x). This gives √3sin(x)cos(x) - sin^2(x) +1=0 4. Using identity 3 above, sin^2(x) → 1 - cos^2(x) simplifying will give √3sin(x)cos(x)+cos^2(x)=0

RP
Answered by Ruadhan P. Maths tutor

9189 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Three forces, (15i + j) N, (5qi – pj) N and (–3pi – qj) N, where p and q are constants, act on a particle. Given that the particle is in equilibrium, find the value of p and the value of q. (Mechanics 1 June 2017)


The equation of a circle is x^2+y^2-6x-4y+4=0. i) Find the radius and centre of the circle. ii) Find the coordinates of the points of intersection with the line y=x+2


Find the stationary points of the function y = (1/3)x^3 + (1/2)x^2 - 6x + 15


If a particle of mass m is launched vertically upwards from the ground with velocity u m/s, how long will it take to return to the ground in terms of m, u and g?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences