Solve Trig Problem w/o Formulas: sin(3pi/2 + a), cos(3pi/2 - x)

[t_n_p]

Homework Statement

If sin x = 0.3, cos a = 0.6 and tan t = 0.7, find the values of:

a) sin (3pi/2 + a)
b) cos (3pi/2 - x)

I can't use use the formulas sin (a + b) = sin(a)cos(b) + cos(a)sin(b)
and cos(a - b) = cos(a)cos(b) - sin(a)sin(b)

because I don't some of the values

But, I'm curious, is there another way to do these questions?

Reason I ask, I was reading a textbook, and these questions were asked, but the above trig identities had not yet been introduced, which led me to think if there were an alternative method. Up to this stage, the textbook had only talked about:

negative angles by symmetry:
cos (−x) = cos x
sin (−x) = −sin x
tan (−x) = −sin x/cos x = −tan x

and the complementary relationships:
sin(pi/2 - x) = cos x
sin(pi/2 + x) = cos x
cos(pi/2 - x) = sin x
cos(pi/2 + x) = -sin x

 

[Calrid]

As long as the trig identities are equivalent you can use whatever one you want (I wouldn't though if the question expects a certain form, use it), that's the beauty of trig identities if you really want to go mental look at all these babies:

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

You're right that was easy.

Incidentally if

sin(pi/2 + x) = cos x

What does Sin (3pi/2+x) = ?

The answer is on that table.

 

[t_n_p]

since cos(3pi/2) = 0 then I guess I can use the trig identity above, my bad!

 

[t_n_p]

sorted.

thanks