Thread: 3 Quick Questions

hey, just need some help on my homework.

First, how would I make a table for a graph of the equation y = -2sin2x. I used 0, pi/2, pi, 3pi/2, 2pi as the x values, but everything equals 0

Second, I have a sheet that says the formulas for double angles, half angles, difference of two angles, etc. nothing with tan though. how would i answer the question "If A and B are acute angles such that cos A = 24/25 and tan B = 15/8, find the exact value of sin(A - B)"


Last question. This one has tan too. the question is "The Double Angle formula for the tangent is: tan2x = (2tanx)/(1-tan^2x). Derive this formula from the formulas for sin2x and cos2x"

(i)to draw the graph, plot all the x intercepts- when 2x=2npi where x is any interger ie ...-3, -2, -1, 0, 1, 2.... and all the maximum, miminium values ie value of 2 when 2x= 2npi -pi where n is any integer and min value -2 when x= 2npi + pi where n is any integer, then join all points

(ii) to find the exact value of sin(A - B) find
sine A with (cos A)^2 + (sin A) ^2 = 1 and cos A = 24/25
sine B with 1 + (cot B)^2 = (cosec B)^2 and tan B = 15/8
cosine B with 1 + (tan B)^2 = (sec B)^2 and tan B = 15/8
** cot B= 1/tanB and cosec B = 1/sinB and sec B = 1/cos B
then use formula
sin(A-B) = sin A cosB -cos A sin B

(iii)tan 2x = sin 2x/ cos 2x
= 2sin x cos x / [(cos x)^2 - (sin x)^2]
divide the denominator and numerator by (cos x)^2 and u will get there

thanks, 1 and 3 helped a lot. im a little confused about 2. the formula is cos A * cos B som A * cos A. what you wrote would just equal 0.

also is there anyway to find the cos and sin of B without using cot or csc?

sorry, lots of typo, coz i am trying to solve my question at the same time, but this edited version should be correct

no problem, still helpful. I got the sin A no problem, but is there any other way to get cos B and sin B without using cot and csc?

unfortunately, i am not an excellent mathematician, at my level i will only be able to tell u the only way to obtain sin x from tan x is through
1+ (cot x)^2 = (cosec x) ^2
and cos x from tan x through
1+ (tan x)^2 = (sec x) ^2