Double angle formulae questions with answers

Questions

1. Question 1Exact values of sin 2A, cos 2A, tan 2A

   Given that $\tan A = \dfrac{15}{8}$ and $A$ is acute, find the exact value of $\cos 2A$.

   Show the worked solution

   1. Step 1: Draw the triangle for tan A: the hypotenuse by Pythagoras' theorem$\text{Hypotenuse} = \sqrt{15^{2} + 8^{2}}$$= 17$
   2. Step 2: Which quadrant is A in? Use ASTC for the signs of sin A and cos A$A$ is in quadrant 1.$\sin A = \dfrac{15}{17}$$\cos A = \dfrac{8}{17}$
   3. Step 3: Use the double angle formula: put the values in, then simplify$\cos 2A = \cos^{2} A - \sin^{2} A$$\cos 2A = \left(\dfrac{8}{17}\right)^{2} - \left(\dfrac{15}{17}\right)^{2}$$= -\dfrac{161}{289}$
2. Question 2Exact values of sin 2A, cos 2A, tan 2A

   Given that $\tan A = \dfrac{5}{12}$ and $A$ is acute, find the exact value of $\cos 2A$.

   Show the worked solution

   1. Step 1: Draw the triangle for tan A: the hypotenuse by Pythagoras' theorem$\text{Hypotenuse} = \sqrt{5^{2} + 12^{2}}$$= 13$
   2. Step 2: Which quadrant is A in? Use ASTC for the signs of sin A and cos A$A$ is in quadrant 1.$\sin A = \dfrac{5}{13}$$\cos A = \dfrac{12}{13}$
   3. Step 3: Use the double angle formula: put the values in, then simplify$\cos 2A = \cos^{2} A - \sin^{2} A$$\cos 2A = \left(\dfrac{12}{13}\right)^{2} - \left(\dfrac{5}{13}\right)^{2}$$= \dfrac{119}{169}$
3. Question 3Equations with sin 2x

   Solve $\sin 2x = 0.8\cos x$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Replace sin 2x by 2 sin x cos x and bring everything to one side$2\sin x\cos x - 0.8\cos x = 0$
   2. Step 2: Factorise: take out the common factor (never divide by it: that loses solutions)$\cos x\Big(2\sin x - 0.8\Big) = 0$
   3. Step 3: From the first factor: Read every angle x in the range off the graph of y = cos x (0, 1 and −1 have no basic angle)$\cos x = 0$$x = 90^\circ$$x = 270^\circ$
   4. Step 4: From the second factor: Basic angle β: use the positive value$\sin x = \dfrac{2}{5}$$\beta = \sin^{-1}\left(0.4000\ldots\right)$$\beta = 23.6^\circ$
   5. Step 5: From the second factor: ASTC (All, Sine, Tan, Cos): in which quadrants does sin x have this sign?$\sin x$ is positive in quadrants 1 and 2.
   6. Step 6: From the second factor: List every angle x in the range, quadrant by quadrant$x = 23.6^\circ$$x = 180^\circ - 23.6^\circ$$= 156.4^\circ$
4. Question 4Equations with sin 2x

   Solve $\sin 2x = \cos x$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Replace sin 2x by 2 sin x cos x and bring everything to one side$2\sin x\cos x - \cos x = 0$
   2. Step 2: Factorise: take out the common factor (never divide by it: that loses solutions)$\cos x\Big(2\sin x - 1\Big) = 0$
   3. Step 3: From the first factor: Read every angle x in the range off the graph of y = cos x (0, 1 and −1 have no basic angle)$\cos x = 0$$x = 90^\circ$$x = 270^\circ$
   4. Step 4: From the second factor: Basic angle β: use the positive value$\sin x = \dfrac{1}{2}$$\beta = \sin^{-1}\left(\dfrac{1}{2}\right)$$= 30^\circ$
   5. Step 5: From the second factor: ASTC (All, Sine, Tan, Cos): in which quadrants does sin x have this sign?$\sin x$ is positive in quadrants 1 and 2.
   6. Step 6: From the second factor: List every angle x in the range, quadrant by quadrant$x = 30^\circ$$x = 180^\circ - 30^\circ$$= 150^\circ$
5. Question 5Equations with cos 2x

   Solve $\cos 2x = \cos x - 1$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Replace cos 2x by the form that leaves only the ratio on the right-hand side$2\cos^{2} x - 1 = \cos x - 1$
   2. Step 2: Bring everything to one side$2\cos^{2} x - \cos x = 0$
   3. Step 3: Factorise as a quadratic in cos x$\cos x(2\cos x - 1) = 0$$\cos x = 0$ or $\cos x = \dfrac{1}{2}$
   4. Step 4: From the first factor: Read every angle x in the range off the graph of y = cos x (0, 1 and −1 have no basic angle)$x = 90^\circ$$x = 270^\circ$
   5. Step 5: From the second factor: Basic angle β: use the positive value$\beta = \cos^{-1}\left(\dfrac{1}{2}\right)$$= 60^\circ$
   6. Step 6: From the second factor: ASTC (All, Sine, Tan, Cos): in which quadrants does cos x have this sign?$\cos x$ is positive in quadrants 1 and 4.
   7. Step 7: From the second factor: List every angle x in the range, quadrant by quadrant$x = 60^\circ$$x = 360^\circ - 60^\circ$$= 300^\circ$
6. Question 6Equations with cos 2x

   Solve $\cos 2x = 2\sin x + 1$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Replace cos 2x by the form that leaves only the ratio on the right-hand side$1 - 2\sin^{2} x = 2\sin x + 1$
   2. Step 2: Bring everything to one side$2\sin^{2} x + 2\sin x = 0$
   3. Step 3: Factorise as a quadratic in sin x$2\sin x(\sin x + 1) = 0$$\sin x = 0$ or $\sin x = -1$
   4. Step 4: From the first factor: Read every angle x in the range off the graph of y = sin x (0, 1 and −1 have no basic angle)$x = 0^\circ$$x = 180^\circ$$x = 360^\circ$
   5. Step 5: From the second factor: Read every angle x in the range off the graph of y = sin x (0, 1 and −1 have no basic angle)$x = 270^\circ$