A-Level Edexcel Maths Pure Modelling with Functions: 12. (a) Prove that $$1

12. (a) Prove that $$1 - ext{cos} 2 heta = an heta rac{ ext{sin} 2 heta}{ ext{sin} 2 heta}, heta eq rac{(2n + 1) ext{π}}{2}, n ext{ } ext{in} ext{ } ext{ℤ}$$ (b) Hence solve, for $$-rac{ ext{π}}{2} < x < rac{ ext{π}}{2},$$ the equation $$( ext{sec} x^2 - 5)(1 - ext{cos} 2x) = 3 an' x ext{ sin } 2x$$ Give any non-exact answer to 3 decimal places where appropriate. - Edexcel - A-Level Maths Pure - Question 14 - 2018 - Paper 2

Question 14

$12.-(a)-Prove-that--$$1----ext{cos}-2-heta-=--an--heta--rac{-ext{sin}-2-heta}{-ext{sin}-2-heta},--heta--eq--rac{(2n-+-1)-ext{π}}{2},-n--ext{-}--ext{in}--ext{-}--ext{ℤ}$$--(b)-Hence-solve,-for--$$--rac{-ext{π}}{2}-<-x-<--rac{-ext{π}}{2},$$--the-equation--$$(-ext{sec}-x^2---5)(1----ext{cos}-2x)-=-3--an'-x--ext{-sin-}-2x$$--Give-any-non-exact-answer-to-3-decimal-places-where-appropriate.-Edexcel-A-Level Maths Pure-Question 14-2018-Paper 2.png$

12. (a) Prove that $$1 - ext{cos} 2 heta = an heta rac{ ext{sin} 2 heta}{ ext{sin} 2 heta}, heta eq rac{(2n + 1) ext{π}}{2}, n ext{ } ext{in} ext{ } ext{... show full transcript

Worked Solution & Example Answer:12. (a) Prove that $$1 - ext{cos} 2 heta = an heta rac{ ext{sin} 2 heta}{ ext{sin} 2 heta}, heta eq rac{(2n + 1) ext{π}}{2}, n ext{ } ext{in} ext{ } ext{ℤ}$$ (b) Hence solve, for $$-rac{ ext{π}}{2} < x < rac{ ext{π}}{2},$$ the equation $$( ext{sec} x^2 - 5)(1 - ext{cos} 2x) = 3 an' x ext{ sin } 2x$$ Give any non-exact answer to 3 decimal places where appropriate. - Edexcel - A-Level Maths Pure - Question 14 - 2018 - Paper 2

Step 1

Prove that $$1 - ext{cos} 2 heta = an heta rac{ ext{sin} 2 heta}{ ext{sin} 2 heta}$$

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Answer

To prove the identity, we start by using the double angle formulas:

The cosine double angle identity states: $ext{cos} 2 heta = 1 - 2 ext{sin}^2 heta$ Substituting this into the left-hand side gives us: $1 - (1 - 2 ext{sin}^2 heta) = 2 ext{sin}^2 heta$ .
For the right-hand side, we use: $an heta = rac{ ext{sin} heta}{ ext{cos} heta}$ . Hence, we have: $an heta ext{ sin} 2 heta = an heta (2 ext{sin} heta ext{cos} heta) = 2 ext{sin}^2 heta$ .
Since both sides equal to $2 ext{sin}^2 heta$ , the proof is complete and verified that: $1 - ext{cos} 2 heta = an heta ext{ sin} 2 heta$ .

Step 2

Hence solve, for $$-rac{ ext{π}}{2} < x < rac{ ext{π}}{2},$$ the equation $$( ext{sec} x^2 - 5)(1 - ext{cos} 2x) = 3 an' x ext{ sin } 2x$$

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Answer

Start with the equation: $( ext{sec} x^2 - 5)(1 - ext{cos} 2x) = 3 an x ext{ sin } 2x$ .
Substitute the known identities: $ext{sec} x = rac{1}{ ext{cos} x}; ext{ and } 1 - ext{cos} 2x = 2 ext{sin}^2 x$
The equation can become: $( rac{1}{ ext{cos}^2 x} - 5)(2 ext{sin}^2 x) = 3 an x(2 ext{sin} x ext{cos} x)$ .
Simplifying gives: $rac{(1 - 5 ext{cos}^2 x)(2 ext{sin}^2 x)}{ ext{cos}^2 x} = 6 ext{sin}^2 x$ .
From this point, arriving at: $1 - 5 ext{cos}^2 x = 6 ext{cos}^2 x$ .
Therefore, we can solve for $ext{cos}^2 x$ :

ightarrow ext{cos}^2 x = rac{1}{11} $leads us to find:$ ext{cos} x = rac{1}{ ext{√11}}$$.

Now finding $ext{sin} x$ gives:

ightarrow ext{sin} x = rac{ ext{√10}}{ ext{√11}}$$.

Finally, substituting $x$ must fall within the range $- rac{ ext{π}}{2} < x < rac{ ext{π}}{2}$ : Applying the arcsin or inverse, the solutions will bring us to approximate values. Finalizing, the answer is to be presented up to three decimal places.

Prove that $$1 - ext{cos} 2 heta = an heta rac{ ext{sin} 2 heta}{ ext{sin} 2 heta}$$

Hence solve, for $$- rac{ ext{π}}{2} < x < rac{ ext{π}}{2},$$ the equation $$( ext{sec} x^2 - 5)(1 - ext{cos} 2x) = 3 an' x ext{ sin } 2x$$

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Hence solve, for $$-rac{ ext{π}}{2} < x < rac{ ext{π}}{2},$$ the equation $$( ext{sec} x^2 - 5)(1 - ext{cos} 2x) = 3 an' x ext{ sin } 2x$$