Given that 'cos A = 3/4' where 270 < A < 360, find the exact value ot sin 2A.
I know sin2A= 2sinAcosA
so, using the triangle, I would get 'sin A = square root of 7/ 4'. And that would be a minus because we're dealing with the 4th quadrant.
so by subbing in 'cos A = 3/4' and 'sin A = square root of 7/ 4' into
'2sinAcosA', I got: 2 * -square root of 7/ 4 * 3/4.
but the mark scheme just has it as '-3*square root of 7 / 8'. I'm confused because shouldnt all that be multiplied by '2' from '2sinAcosA'
by the way for equations involving modulus, for example |2x+3|. Do I treat it as '-(2x+3)'
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C3 questions .....need help watch
- Thread Starter
- 17-01-2010 18:51
- 17-01-2010 18:57
Sorry not to answer the first bit but for |2x+3| you tread it as +/- (2x+3)
e.g. If |2x+3|=12
Either 12 = (2x+3) -> x = 4.5
Or 12 = -(2x+3) -> x = -7.5Last edited by Schismist; 17-01-2010 at 18:59.
- 17-01-2010 19:04
what board is this by any chance, i havent seen these type questions in ocr mei.
- 17-01-2010 19:08
- 17-01-2010 19:09
- 17-01-2010 19:12
For equation involving a modulus, the value of x is always such that the modulus is positive. For example, if you had |2x + 3|, then at values of x less than -1.5, x becomes positive.