What is the GCF of 44j^5k^4 and 121j^2k^6?
2 answers:
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Answer:
Step-by-step explanation:
<h3>Given expression</h3>sin²5° + sin²10° + sin²15° +... sin²85° + sin²90°
<h3>Concept:</h3>sin²x + cos²x = 1
sin(x) = cos (90 - x)
<u />
<u>There are in total these terms:</u>
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
<u>In total, there are 18 terms, and the first one matches with the second to the last one:</u>
5 -- 85
10 -- 80
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40 -- 50
<u />
<u>There are 2 terms left over:</u>
sin²45 and sin²90
<h3>Convert the first half of the sine terms (sin²5 - sin²40) to the cosine terms</h3>sin²5 = cos² (90 - 5) = cos²85
sin²10 = cos² (90 - 10) = cos²80
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sin²40 = cos² (90 - 40) = cos²50
<h3>Simplify the 16 grouped terms </h3><em>i.e. sin²85 and cos²85</em>
Using the concept of sin²x + cos²x = 1
sin²85 + cos²85 = 1
sin²80 + cos²80 = 1
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sin²50 + cos²50 = 1
Total = (16/2) × 1 = 8 × 1 = 8
<h3>Evaluate the 2 terms that are left over</h3>sin²45 = (sin45) (sin45) = (√2 / 2) (√2 / 2) = 1/2
sin²90 = (sin90) (sin 90) = (1) (1) = 1
<h3>Add all the terms together</h3>Hope this helps!! :)
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