a is in quadrant 2
b is in quadrant 4
5. Find cos(a+b)if cos a=4/5 & sin b=12/13
a is in quadrant 1
b is in quadrant 1
6. Find tan(a-b)if sin a=-5/13 & tan b=3/4
a is in quadrant 4
b is in quadrant 3
7. Find sin2theta if sin theta=2/3
theta is in quadrant 2
8. If sin theta= -3/5 & theta is in quadrant 3, find cos theta/2
9. Find tan 2theta if tan theta=5/12 and theta is in quadrant 3
omighosh------that's a lot many questions. i'll try to ans as many as i can-----till my patience doesn't wear out!!
square both sides
(sin a)^2=16/25 ................. 1
we know, (sin a)^2=1-(cos a)^2
subs, in 1
1 - (cos a)^2=16/25
cos a=+3/25 or -3/25
since a is in quadrant 2,
cos a is negative . hence cos a is -3/25 ............... 2
similarly, cos b is +12/13 or -12/13
but b is in the fourth quadrant , where sin is -ve, but cos is +ve
hence cos b is +12/13
subs these vals in the given expansion of question
the answer will be : (4/5)(-3/5)-(-5/13)(12/13)
which equals -528/4225 in its simplest form.
5.you can find sin a from cos a by the same method give above . since a is in quadrant 1, sin a is +ve. i.e. +3/5
and cos b also is +ve(1st quadrant) and thus equal to 5/13.
cos(a+b)=cos a cos b - sin a sin b
which, after subs vals , you get 17/65 in its simplest form
6. in the 6th question the same basics are applied . from sin(a) find cos (a). (you know how?!). the ans will be positive(look at the quadrant) you will get cos a=12/13.
now, tan x=sin x/cos x
hence tan a= - 5/12
tan(a-b)=(tan a-tan b)/(1 +tan(a)tan(b))
subs values ...please solve it yourself?
7. you know, sin 2x=2 sin(x)cos(x)
find cos(x) and subs
8. now, use cos x=2.(cos (x/2))^2 - 1 ................ take theta as x
i think we have done enough now for you to know how to find cos x . it will be -ve(have a look at the quadrant?)
now put the value in the eqn and solve for cos(x/2)
9.tan 2x= (2tan x)/(1 - (tan x)^2)
if you subs value of tan x on the right hand side, you'll get tan 2x ................................. theta is taken as x.
and that finishes it .
tan(a-b)=tana-tanb/1+tana*tanb (the part before the slash is over the part after the slash)
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