Advertisements
Advertisements
Question
Evaluate cos 48° − sin 42°
Solution
We have to find cos 48° − sin 42°
Since `cos (90^@ - theta) = sin theta` So
`cos 48^@ - sin 42^@ = cos (90^@ - 42^@) - sin 42^@`
`= sin 42^@ - sin 42^@`
= 0
So value of `cos 48^@ - sin 42^@` is 0
APPEARS IN
RELATED QUESTIONS
If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ – 1
`(1- tan^2 45°)/(1+tan^2 45°)` = ______
Evaluate the following :
`((sin 27^@)/(cos 63^@))^2 - (cos 63^@/sin 27^@)^2`
Evaluate: `(2sin 68)/cos 22 - (2 cot 15^@)/(5 tan 75^@) - (8 tan 45^@ tan 20^@ tan 40^@ tan 50^@ tan 70^@)/5`
Evaluate:
`(cos3"A" – 2cos4"A")/(sin3"A" + 2sin4"A")` , when A = 15°
If A =30o, then prove that :
sin 2A = 2sin A cos A = `(2 tan"A")/(1 + tan^2"A")`
prove that:
tan (2 x 30°) = `(2 tan 30°)/(1– tan^2 30°)`
Given A = 60° and B = 30°,
prove that : cos (A + B) = cos A cos B - sin A sin B
Find the value of x in the following: tan x = sin45° cos45° + sin30°
Find the value of the following:
`(tan45^circ)/("cosec"30^circ) + (sec60^circ)/(cot45^circ) - (5sin90^circ)/(2cos0^circ)`
