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Sum and difference identities Quiz 2

mathematics

 Solve the equation ${{\sin }^{2}}x+{{\sin }^{2}}2x=1$

 Apply the formula of $\sin 2x=2\sin x\cos x$ and ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ to solve the trigonometric equation.

  ${{\sin }^{2}}x+{{\sin }^{2}}2x=1 \\ \Rightarrow {{\sin }^{2}}x+{{\left( 2\operatorname{sinxcosx} \right)}^{2}}=1 \\ \Rightarrow {{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x=1 \\ \Rightarrow {{\sin }^{2}}x+4{{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)=1 \\ \Rightarrow {{\sin }^{2}}x+4{{\sin }^{2}}x-4{{\sin }^{4}}x=1 \\ \Rightarrow 5{{\sin }^{2}}x-4{{\sin }^{4}}=1 \\ \text{Hence }5{{\sin }^{2}}x-4{{\sin }^{4}}x=1$

 Solve ${{\cos }^{2}}2x+{{\cos }^{2}}4x=1$

 Apply the formula of $\cos 2x=2{{\cos }^{2}}x-1$ and $\sec x=\dfrac{1}{\cos x}$ to solve the given trigonometric equation.

  ${{\cos }^{2}}2x+{{\cos }^{2}}4x=1 \\ \Rightarrow {{\cos }^{2}}2x+{{\left( 2{{\cos }^{2}}2x-1 \right)}^{2}}=1 \\ \Rightarrow {{\cos }^{2}}2x+4{{\cos }^{4}}2x+1-4{{\cos }^{2}}2x=1 \\ \Rightarrow {{\cos }^{2}}2x+4{{\cos }^{4}}2x-4{{\cos }^{2}}2x=0 \\ \Rightarrow 4{{\cos }^{4}}2x-3{{\cos }^{2}}2x=0 \\ \Rightarrow \dfrac{4}{{{\sec }^{4}}2x}-\dfrac{3}{{{\sec }^{2}}2x}=0 \\ \Rightarrow \dfrac{4-3{{\sec }^{2}}2x}{{{\sec }^{4}}2x}=0$

 Solve ${{\cos }^{2}}x-2{{\cos }^{2}}\dfrac{x}{2}=0$

 Apply the formula of $\cos 2x=2{{\cos }^{2}}x-1$ and $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the given trigonometric equation.

 ${{\cos }^{2}}x-2{{\cos }^{2}}\dfrac{x}{2}=0 \\ \Rightarrow {{\left( 2{{\cos }^{2}}\dfrac{x}{2}-1 \right)}^{2}}-2{{\cos }^{2}}\dfrac{x}{2}=0 \\ \Rightarrow 4{{\cos }^{4}}\dfrac{x}{2}+1-4{{\cos }^{2}}\dfrac{x}{2}-2{{\cos }^{2}}\dfrac{x}{2}=0 \\ \Rightarrow 4{{\cos }^{4}}\dfrac{x}{2}-6{{\cos }^{2}}\dfrac{x}{2}+1=0 \\ a=4$, $b=-6$, $c=1 \\ \Rightarrow {{\cos }^{2}}\dfrac{x}{2}=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 4 \right)\left( 1 \right)}}{2\left( 4 \right)} \\ =\dfrac{6\pm \sqrt{36-16}}{8} \\ =\dfrac{6\pm \sqrt{20}}{8} \\ =\dfrac{6\pm 2\sqrt{5}}{8} \\ =\dfrac{3\pm \sqrt{5}}{4} \\ {{\cos }^{2}}\dfrac{x}{2}=\dfrac{3+\sqrt{5}}{4}$ and ${{\cos }^{2}}\dfrac{x}{2}=\dfrac{3-\sqrt{5}}{4}$

 Apply the formula of ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ and $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the given trigonometric equation.

 ${{\cos }^{2}}x-{{\tan }^{2}}x=0 \\ \Rightarrow \dfrac{1}{{{\sec }^{2}}x}-{{\tan }^{2}}x=0 \\ \Rightarrow \dfrac{1}{1+{{\tan }^{2}}x}-{{\tan }^{2}}x=0 \\ \Rightarrow 1-{{\tan }^{2}}x\left( 1+{{\tan }^{2}}x \right)=0 \\ \Rightarrow 1-{{\tan }^{2}}x-{{\tan }^{4}}x=0 \\ \Rightarrow {{\tan }^{4}}x+{{\tan }^{2}}x-1=0 \\ \Rightarrow {{\tan }^{2}}x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\ \Rightarrow \dfrac{-1\pm \sqrt{1+4}}{2}=\dfrac{-1\pm \sqrt{5}}{2} \\ {{\tan }^{2}}x=\dfrac{-1+\sqrt{5}}{2}\text{ and }\dfrac{-1-\sqrt{5}}{2}$

 Choose the correct answer $\tan 3x\tan x\tan 4x=?$

 Apply the formula of $\tan \left( A+B \right)=\dfrac{\operatorname{tanA}+tanB}{1-\operatorname{tanAtanB}}$ to solve the given trigonometric equation.

 $\Rightarrow \tan 4x=\tan \left( x+3x \right) \\ \Rightarrow \tan 4x=\dfrac{\tan x+\tan 3x}{1-\tan x\tan 3x} \\ \Rightarrow \tan 4x\left( 1-\tan x\tan 3x \right)=\tan x+\tan 3x \\ \Rightarrow \tan 4x-\tan x\tan 3x\tan 4x=\tan x+\tan 3x \\ \Rightarrow \tan 4x-\tan x-\tan 3x=\tan x\tan 3x\tan 4x \\ \text{Hence }\tan x\tan 3x\tan 4x=\tan 4x-\tan x-\tan 3x$