Question

4 (sin 4 30 cos 4 60 )- 3 (cos 2 45 sin 2 90) - 4sin2 45
irrevelent answer will be reprted 100%

Answer 1

Answer:

Value\: of \: 4(sin^{4}30+cos^{4}60)-3(cos^{2}45-sin^{2}90)Valueof4(sin430+cos460)−3(cos245−sin290)

= 2=2

Explanation:

Value\: of \: 4(sin^{4}30+cos^{4}60)-3(cos^{2}45-sin^{2}90)Valueof4(sin430+cos460)−3(cos245−sin290)

=[4(\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{4}]-3[\left(\frac{1}{\sqrt{2}}\right)^{2}-1^{2}][4((21)4+(21)4]−3[(21)2−12]

=

= 4\times \frac{2}{16}-3(\frac{(1-2)}{2})4×162−3(2(1−2))

= \frac{1}{2}-3\left(\frac{-1}{2}\right)21−3(2−1)

=\frac{1}{2}+\frac{3}{2}=21+23

=\frac{4}{2}=24

= 2=2

Therefore,

Value\: of \: 4(sin^{4}30+cos^{4}60)-3(cos^{2}45-sin^{2}90)Valueof4(sin430+cos460)−3(cos245−sin290)

= 2=2

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Answer 2

Step-by-step explanation:

4(sin430∘+cos260∘)−3(cos245∘−sin290∘)−sin260∘

=4((21)4+(21)2)−3((21)2−12)−(23)2

=4((161)+(41))−3((21)−1)−(43)

=4((161)+(