Half Angle And Double Angle Formulas Math

Scroll down the page for more examples and solutions on how to use the half angle identities and double angle identities.

Half angle and double angle formulas math. Given the tangent of an angle and the quadrant in which it is located use the double angle formulas to find the exact value. Scroll down the page for more examples and solutions of how to use derive and proof the double angle formulas and half angle formulas. Given cos a 2 3 in quadrant ii determine cos a 2 sin a 2 and tan a 2. The following figure gives the double angle formulas and half angle formulas.

Notice that this formula is labeled 2 prime. From the double angle formulas one may generate easily the half angle formulas in particular we have example. Check the identities answer. If we replace with the half angle formula for sine is found by simplifying the equation and solving for note that the half angle formulas are preceded by a sign.

Then using the above formulas we get since then is a positive number. Determine the exact value of sin π 8 2. If you multiply top and bottom by 1 cos b instead of 1 cos b you get another form of the half angle tangent formula. Find the exact value for sin 105 using the half angle identity.

Half angle formulas multiple angle formulas powers of. The double angle formulas are summarized as follows. Tan b 2 sin b 2 cos b 2 1 cos b 1 cos b tan b 2 1 cos b 1 cos b 1 cos b tan b 2 1 cos b 1 cos b. Determine the exact value of cos 105 3.

Tan 2a 2 tan a 1 tan 2 a. Questions for problems 1 6 sin theta frac 3 5 and 90 circ lt theta lt180 circ find each value. Use the half angle formulas to find answer. This is the half angle formula for the cosine.

Again whether we call the argument θ or does not matter. The next set of identities is the set of half angle formulas which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. The sign will depend on the quadrant of the half angle. Sin 2a 2 sin a cos a.

Cos 2a cos 2 a sin 2 a 1 2 sin 2 a 2 cos 2 a 1. Sin 2θ 2sinθcosθ cos 2θ cos2θ sin2θ 1 2sin2θ 2cos2θ 1 tan 2θ 2tanθ 1 tan2θ. Therefore we have same arguments lead to example. Given find a sin 2θ b solution.

The sign of the two preceding functions depends on the quadrant in which the resulting angle is located. A sin 2θ 2 sin θ cos θ.