We now proceed to derive two other related formulas that can be used when proving trigonometric identities.
It is suggested that you remember how to find the identities, rather than try to memorise each one.
Think of it as a reflection; like looking in a mirror.
An example of a trig identity is \(\displaystyle \csc (x)=\frac\); for any value of \(x\), this equation is true.
\(\displaystyle \begin&=\frac-\frac\\\tan \left( \right)&=\frac\,\,\,\,\,\surd \end\) Note: Start from the right side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos.
\(\displaystyle \begin\frac&=\\frac&=\cos \theta \sin \theta \,\,\,\,\,\surd \end\) Note: We knew to use \(\displaystyle \theta -\theta \) and difference of squares for \(\cos 2\theta \) since the denominator contains both cos and sin.Trig identities are sort of like puzzles since you have to “play” with them to get what you want.You will also have to do some memorizing for these, since most of them aren’t really obvious.\(\displaystyle \begin\cos \left( \right)\cos \left( \right)&=\frac\\left( \right)\left( \right)&=\frac\\left( \right)\left( \right)&=\frac\\fracx-\fracx&=\frac\\frac\left( \right)-\fracx&=\frac\3-3x-x&=2\-4x&=-1\x&=\frac\\sin x&=\pm \frac\end\) \(\displaystyle x=\left\) We use double angle and half angle identities the same way we used sum and difference identities when we need to split up the angle to make it easier to find the values (for example, to find values on the unit circle).We also use the identities in conjunction with other identities to prove and solve trig problems.For the \(\displaystyle \sin \left( \right)\) and \(\displaystyle \cos \left( \right)\), they both contain a cos, and the cos half angle has the plus sign inside the radical.The tricky thing is that the sign of the whole identity depends on the quadrant where the angle \(\displaystyle \boldsymbol\) is (not angle \(A\), but \(\displaystyle \frac\)).For \(\displaystyle \tan \left( \right)\), you’ll want to memorize both parts of the identity, but you can derive the second part by multiplying the first by \(\displaystyle \frac\) (conjugate of denominator).\(\require \displaystyle \begin&=\frac\&=\frac\cdot \frac\&=\sin x\cos x\&=\sin x\cos x\cdot \frac\\frac&=\frac\,\,\,\,\,\surd \end\) Note: To get \(\displaystyle \sin x\cos x\) into the right form for the identity \(\displaystyle (2\sin x\cos x)\), we had to multiply by \(\displaystyle \frac\).Here are the double angle and half angle identities, and tricks to help you memorize them: ” out and put it in front of a sin and cos.\(\cos \left( \right)\) is a little more complicated, especially since it can be written in three ways.