Mathematics
01.10.2022 10:43
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Prove that cos²(A-B)-sin²(A+B)=cos2Acos2B

Prove that cos²(A-B)-sin²(A+B)=cos2Acos2B
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QueensQueens
QueensQueens
4,6(72 marks)

see below

Step-by-step explanation:

Prove : cos²(A-B)-sin²(A+B)=cos2Acos2B

LHS,

cos²(A-B) - sin²(A+B)     (expand using double angle identities)

= \frac{1 + cos [2(A-B)]}{2}  - \frac{1 - cos [2(A+B)]}{2}

= \frac{1}{2}  + \frac{1}{2} cos[2(A-B)] - \frac{1}{2} +  \frac{1}{2} cos[2(A+B)]

= \frac{1}{2} cos[2(A-B)] +  \frac{1}{2} cos[2(A+B)]

= \frac{1}{2} cos(2A-2B) +  \frac{1}{2} cos(2A+2B)      (expand using sum/difference identities)

= \frac{1}{2} (cos2Acos2B + sin2Asin2B ) + \frac{1}{2} (cos2Acos2B - sin2Asin2B )

= \frac{1}{2} cos2Acos2B + \frac{1}{2}sin2Asin2B  + \frac{1}{2} cos2Acos2B - \frac{1}{2}sin2Asin2B

= \frac{1}{2} cos2Acos2B  + \frac{1}{2} cos2Acos2B

= cos2Acos2B  (= RHS , Proven)

tangia
tangia
4,5(59 marks)

If you start at (0, 0) and move right 3 and up 6, you arrive at (3, 6). Then, starting at (3, 6) if you move right 5 and up -5 (i.e. down 5), you end up at (8, 1). The resultant is the vector from your original starting point to your final landing spot. So the vector from (0, 0) to (8, 1), which is simply <8, 1>.

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