Rhys_M
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When solving trig equations with double angles, when do you have to add on the adjusted period to find more solutions and when don't you?
Eg:
For 1+cos2θ=7 cos²θ-1 (which became 5cos²θ=1), I got θ=63.435 but then do 180-63.... to get another solution of 116.6˚

But for tanθtan2θ=3 , which because tan²θ=3/5 , I did this (halved tan period to get 90˚, so did 90-37.76 and 90+37.67), but these aren't solutions: to get the other solution you do 180-37.8 = 142.2˚.

So what are the rules for finding other solutions when double angles are involved?
Thank you
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mqb2766
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You're really asking about squared trig functions, not double angles (though they are obviously related). So the partial solution to
cos^2(x) = 1/5
is
cos(x) = +/- 1/sqrt(5)
which gives the reflection in the y axis (90 degrees or pi/2).

Use the sinusoidal curves and/or the cast diagram to identify the +/- trig values. Its the normal rules, nothing special.
Last edited by mqb2766; 4 weeks ago
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Rhys_M
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(Original post by mqb2766)
You're really asking about squared trig functions, not double angles (though they are obviously related). So the partial solution to
cos^2(x) = 1/5
is
cos(x) = +/- 1/sqrt(5)
which gives the reflection in the y axis (90 degrees or pi/2).

Use the sinusoidal curves and/or the cast diagram to identify the +/- trig values. Its the normal rules, nothing special.
Ohhh yea of course - i was focused on using the double angle formulae i just learnt so forgot that basic thing - thanks
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mqb2766
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In a sense you are right, if you use the double angle directly (rather than converting to a squared single angle) you would need to be careful about careflly writing the multiple solutions. But thats not what you're doing here.
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Rhys_M
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(Original post by mqb2766)
In a sense you are right, if you use the double angle directly (rather than converting to a squared single angle) you would need to be careful about careflly writing the multiple solutions. But thats not what you're doing here.
Alright, thank you
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