MENU
Educational Resources
Exam Papers
Form 1 Videos
Form 2 Videos
Form 3 Videos
Form 4 Videos
Grade 4 Videos
Grade 5 Videos
Grade 6 Videos
Grade 7 Videos
Class 8 Videos
Form 1 Exams
Form 2 Exams
Form 3 Exams
Form 4 Exams
KCSE Videos
Class 8 Exams
Grade 5 Exams
Grade 4 Exams
Grade 3 Exams
Grade 2 Exams
Grade 1 Exams
Online Tests
Online Tuition
Sign In
Join
Get access to thousands of educational resources
Get premium membership
and access revision papers with marking schemes, video lessons and live classes.
OR
Processing. Please wait.
Form 4 mathematics trigonometry III questions and answers
Determine the amplitude and the period for the graph of y =6 sin #{ x/2 − 90}^0#
(2m 33s)
1915 Views
SHARE
|
« Previous
Next »
1.
Solve the equation:Sin (#1/2x-30^0#) = cos x for 0 < x < #90^0#
2.
Solve for x: 4 sin #(x +20)^0# =3 for #0^0# = x = #360^0#
3.
Solve the equation;sin #5/2theta =1/2#for #0^0 = theta = 180^0#
4.
Solve for #theta# in the equation: #Sin (2theta – 10^0) #= -0.5 for #0^0 = theta = 360^0#
5.
Determine the amplitude and the period for the graph of y =6 sin #{ x/2 - 90}^0#
6.
Find all the values of #theta# between #0^0# and #360^0# satisfying the equation 5 sin #theta# = -4.
7.
Solve the equation;cos #(3theta+120^0)=frac{sqrt3}{2}# For # 0^0 = theta = 180^0#
8.
Solve the equation: #8s^2 + 2s -3 =0# Hence solve the equation: #8sin^2theta +2sintheta-3=0# For #0^0 = theta = 180^0#
9.
Solve the equation: #2sin^2#(x -#30^0#) = cos #60^0# For -#180^0 = x= 180^0#
10.
Given that sin (x +#30^0#) = cos 2x for #0^0#= x= #90^0# find the value of x. Hence find the value of #cos^2#3x.
11.
Solve the equation: #4Sin^2theta# + #4 Cos theta# = 5 for #0^0 = theta = 360^0#.Give the answer in degrees.
12.
Solve the equation: #3tan^2#x – 4 tan x – 4 =0. For #0^0 = x = 180^0#.
13.
Given that #x^0# is an angle in the first quadrant such that 8#sin^2x# +2cosx -5 =0 find; a) cos x b).tan x.
14.
Given that Cos 2x = 0.8070, find x when #0^0 = x = 180^0#
15.
Solve the equation 3cos x =#2sin^2x# where #0^0= x = 360^0#
16.
Solve the equation: 2#cos2theta# = 1 For #0^0= theta = 360^0#
17.
Solve the equation: Sin #(3x+30)^0# =#frac{sqrt3}{ 2}# ,for For #0^0 = x = 90^0#
18.
Solve:4 – 4 #cos^2alpha# = 4sin#alpha# -1 For #0^0 = alpha = 360^0#
19.
Solve the equation: Sin#(2t +10)^0# =0.5 for #0^0= t = 180^0#
20.
Given that sin#(x+20)^0# = -0.7660, find x to the nearest degree, for #0^0= x = 360^0#
×
Share Content Via:
Facebook
Twitter
WhatsApp
Close
Get premium membership