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Well got some trigonometric task and need help in solving them....
If theta lies in Q3 then then cos theta/2 + sin theta/2 is +ve.. How to prove?
(sorry i am using old nokia 2700 so i can't use theta sign)
Math may be a bitter truth but truth is always useful.....
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hi Geeky,
Welcome to the forum.
There is something missing from your first post. Here is the graph of cos theta/2 + sin theta/2
As you can see, the function is another sine curve and is negative for half the time.
So what else does the question say?
Here is the proof that it is a sine curve:
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thnx but i didn't undestand the step 2(how it came from step 1) of ur prove i mean to say that did u omit any step...
Math may be a bitter truth but truth is always useful.....
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hi Geeky,
If you've met this technique before then all the steps are there. But, as you are new to it, I'll give more details.
Have you met the formula:
If this is new to you, I'll have to explain that as well, but let us just take it as correct for now.**
If you have an expression like this:
where p and q are numbers, you can always use this method to change the expression into a single sine expression.
What is needed is to make p into cos(B) and q into sin(B), but that is not likely to be possible. They would have to be fractions for a start, and even if both p and q were fractions they would have to be the right fractions, so that inverse cos(p) and inverse sin(q) give the same angle B.
But there is a simple way to make p and q do what we want.
Calculate R = √ (p^2 + q^2) and then multiply and divide by R like this:
Because of the way that R was calculated it acts as the hypotenuse where p is the adjacent and q is the opposite, so this creates two fractions with the right property:
Because of the general properties of trig. functions this even works when p or q are negative. So we have
With your question p = q = 1, so that makes R = √2
Now sin(π/4) = cos(π/4) = 1/√2, so that's where step 2 comes from.
**If any of this is new to you, post back for an explanation. It would help to have an idea about your current level of maths.
I am using radians ie. pi/4 rather than 45 degrees .
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I am known to other but not this one:
R{p/Rsin(A).q/Rcos(B)}=R{cosB.sinA+sinB.cosA}
Math may be a bitter truth but truth is always useful.....
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Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thnx buddy
Math may be a bitter truth but truth is always useful.....
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