h (1- sin 2 q)(1+ tan 2 q) j sec 4 q - 2sec 2 q tan 2 q + tan 4 q 2 cosec x =
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1 Trigonometric Functions 6D a Use + tan q sec q with q replaced with q + tan q ( ) sec ( q ) b (secq -)(secq +) sec q - (+ tan q) - tan q c tan q(cosec q -) ( ) tan q (+ cot q) - tan q cot q tan q d (sec q -)cotq tan q cotq tan q tan q tanq tanq e (cosec q - cot q) ( ) (+ cot q) - cot q f - tan q + sec q - tan q + (+ tan q) g - tan q ++ tan q 3 tanq secq + tan q tanq secq sec q tanq secq tanq cosq sinq cosq cosq sinq (multiply out) h (- sin q)(+ tan q) i cos q sec q cos q cosecq cotq + cot q cos q cosecq cotq cosec q cosecq cotq sinq cosq sinq cosq j sec 4 q - sec q tan q + tan 4 q (sec q - tan q) (factorise) ( ) (+ tan q) - tan q k 4cosec q + 4cosec q cot q 4cosec q(+ cot q) 4cosec q cosec q 4cosec 4 q k cosec x cosec x Þ cosec x k Þ + cot x k Þ cot x k - Þ cot x ± k - Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
2 3 a cotq 3 90 <q <80 Þ cot q 3 Þ + cot q Þ cosec q 4 Þ sin q 4 5 cosq 4 5, q reflex As cosq is positive and q reflex, q is in the 4th quadrant. Use right-angled triangle where cosj 4 5 Þ sinq (as q is in nd quadrant, sinq is positive) b Using sin q + cos q Þ cos q - sin q Þ cosq (as q is in nd quadrant, cosq is negative) 4 tanq <q < 70 Using Pythagoras' theorem, 5 x + 4 Þ x Þ x 7 So tanj 7 4 and sinj 7 5 As q is in the 4th quadrant, both tanq and sinq are negative Draw a right-angled triangle where tanj 3 4 a tanq b cosecq sinq Using Pythagoras' theorem, x 5 So cosj 4 5 and sinj 3 5 As q is in the 3rd quadrant, both sinq and cosq are negative. a secq cosq - cosj a LHS º sec 4 q - tan 4 q º (sec q - tan q)(sec q + tan q) (difference of two squares) º ()(sec q + tan q) (as + tan q º sec q Þ sec q - tan q º ) º sec q + tan q º RHS b LHS º cosec x - sin x º (+ cot x) - (- cos x) º + cot x -+ cos x º cot x + cos x º RHS b cosq -cosj c sinq -sinj Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
3 6 c LHS º sec A(cot A- cos A) æ cos A º cos A sin A - ö cos A ø º sin A -º cosec A - (use + cot q cosec q) º + cot A - º cot A º RHS d RHS º (sec q -)(- sin q) º tan q cos q (use + tan q º sec q and cos q + sin q º ) º sin q cos q cos q º sin q º - cos q º LHS e LHS º - tan A + tan A º - tan A sec A º sec A (- tan A) f æ º cos A - sin A ö cos A ø º cos A - sin A º (- sin A) - sin A º - sin A º RHS RHS º sec q cosec q º sec q(+ cot q) º sec q + cos q cos q sin q º sec q + sin q º sec q + cosec q º LHS g LHS º cosec Asec A º cosec A(+ tan A) º cosec A + sin A sin A cos A º cosec A + sin A cos A º cosec A + sin A cos A cos A º cosec A + tan Asec A º RHS h LHS º (secq - sinq)(secq + sinq) º sec q - sin q º (+ tan q) - (- cos q) º + tan q -+ cos q º tan q + cos q º RHS 7 3tan q + 4sec q 5 Þ 3tan q + 4(+ tan q) 5 Þ 3tan q tan q 5 Þ 7 tan q Þ tan q 7 Þ cot q 7 Þ cosec q - 7 Þ cosec q 8 Þ sin q 8 As q is obtuse (in the nd quadrant), sinq is positive. So sinq 8 4 Alternatively LHS º sec q + cosec q º cos q + sin q º sin q + cos q cos q sin q º cos q sin q º sec q cosec q º RHS Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 3
4 8 a sec q 3tanq 0 q 360 Þ + tan q 3tanq Þ tan q - 3tanq + 0 tanq 3± 5 (equation does not factorise) For tanq 3+ 5, calculator value is 69. (3 s.f.) b tan q - secq p q p Þ (sec q -) - secq + 0 Þ sec q - secq 0 Þ secq(secq - ) 0 Þ secq (as secq cannot be 0) Þ cosq Þ q - p 3, p 3 Solutions are 69., 49 For tanq 3-5, calculator value is 0.9 (3 s.f.) c cosec q + 3cotq -80 q 80 Þ (+ cot q) + 3cotq Þ cot q - 3cotq + 0 Þ (cotq -)(cotq - ) 0 Þ cotq or cotq Þ tanq or tanq Solutions are 0.9, 0 Set of solutions: 0.9, 69., 0, 49 (3 s.f.) tanq Þ q -35, 45 tanq Þ q -53, 6.6 (3 s.f.) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4
5 8 d cotq - cosec q 0 q p Þ cotq - (+ cot q) Þ cotq -cot q Þ cot q + cotq 0 Þ cotq(cotq +) 0 Þ cotq 0 or cotq - For cotq 0 refer to graph: q p, 3p For cotq -, tanq - f (secq - cosq) tanq - sin q 0 q p Þ sec q - secq cosq + cos q tanq - sin q Þ sec q - + cos q tanq - sin q æ secq cosq cosq cosq ö ø Þ (+ tan q) - + (cos q + sin q) tanq Þ + tan q - + tanq Þ tan q - tanq 0 Þ tanq(tanq -) 0 Þ tanq 0 or tanq tanq 0 Þ q 0, p tanq Þ q p 4 Set of solutions: 0, p 4, p So q 3p 4, 7p 4 Set of solutions: p, 3p 4, 3p, 7p 4 e 3sec q tan q 0 q 360 g tan q secq - 0 q 80 Þ sec q - secq - Þ sec q - secq 0 Þ secq(secq -) 0 Þ secq 0 (not possible) or secq Þ cosq 0 q 360 Refer to graph of y cosq Þ q 0, 360 Þ q 0, 80 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5
6 8 h sec q - (+ 3)tanq + 3 for 0 q p Þ (+ tan q) - (+ 3)tanq + 3 Þ tan q - (+ 3)tanq Þ (tanq - 3)(tanq -) 0 Þ tanq 3 or tanq c Solutions of tan k seck, 0 k 360 are solutions of cos k - Calculator solution is 65.5 ( d.p.) Þ k 65.5, , 94.5 ( d.p.) 0 a As a 4sec x Þ sec x a 4 Þ cos x 4 a As cos x b Þ b 4 a b c cot x First answer for tanq 3 is p 3 Second solution is p + p 3 4p 3 First answer for tanq is p 4 Second solution is p + p 4 5p 4 Set of solutions: p 4, p 3, 5p 4, 4p 3 9 a tan k seck Þ (sec k -) sec k Þ sec k - sec k - 0 Þ sec k ± 8 ± ± As sec k has no values between - and sec k + b cos k ( )( -) Þ c cot x Þ c tan x Þ c sec x - (use + tan x º sec x) Þ c a 6 - æ sec x a ö 4ø Þ 6 a c -6c (multiply by 6c ) Þ c (a -6) 6 Þ c 6 a -6 a x secq + tanq x secq + tanq secq - tanq (secq + tanq)(secq - tanq) secq - tanq sec q - tan q secq - tanq (as + tan q º sec q Þ sec q - tan q º ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
7 b x + secq + tanq + secq - tanq x secq æ Þ x + ö xø 4sec q Þ x + x x + x 4sec q Þ x + x + 4sec q sec q - tan q p Þ (+ tan q) - tan q p Þ + tan q - tan q p Þ tan q p - Þ cot q p - ( p ¹ ) cosec q + cot q + p - ( p - ) + p - p - p -, p ¹ Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
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