evaluate the following `sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°` - trigonometric ratios
evaluate the following `sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°`
`sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°` ....(i) by trigonometric ratios we have `sin 30^@ = 1/2` `cos 45^@ = 1/sqrt2` `tan 30^2 = 1/sqrt3` `sin 90^@ = 1 cos 90^@ = 0 cos 0^@ = 1` by substituting above values in (i), we get `[1/2]^2 . [1/sqrt2]^2 + 4[1/sqrt3]^2 + 1/2[1]^2 - 2[0]^2 + 1/24 [1]^2` `1/4.1/2 + 4/ 1/3 + 1/2 - 0 + 1/24` `1/8 + 4/3 + 1/2 + 1/24 = 48/24 = 2`
sin(α β) = sinαcosβ cosαsinβ 
sin(α-β) = sinαcosβ - cosαsinβ


cos(α β) = cosαcosβ - sinαsinβ
cos(α-β) = cosαcosβ sinαsinβ


tan(α β) = tanα tanβ / 1-tanα tanβ
tan(α-β) = tanα-tanβ / 1 tanα tanβ


sin2θ = 2sinθcosθ = 2tanθ /1 tanθtanθ


cos2θ = cos平方θ−sin平方θ = 2cos平方θ−1 = 1−2sin平方θ


△abc=海龍公式 √s(s-a)(s-b)(s-c) 
[s=a b c/2 ，就是周長/2 ]



△abc=1/2 ａ.b*sinc = axbxc / 4r = rxs 

 [r=外接圓半徑，r=內接圓半徑]




應該是這樣吧!希望對你有幫助。 dear student,
1+ tan2x= 1+sin2x/cos2x
find common denominator which is cos2x
cos2x/cos2x +
sin2x/cos2x
(cos2x + sin2x)/cos2x
1+cot2x = 1+ cos2x/sin2x
common denominator is sin2x
sin2x/sin2x +
cos2x/sin2x
( sin2x + cos2x)/sin2x
1/(cosx2+ sin2x)/cos2x + 1/(
sin2x + cos2x)/sin2x
multiply the numerator and denominator by the reciprocal of the
bottom fractions to get
cos2x/(cosx2+ sin2x )+
sin2x/( cosx2+ sin2x)
since they now have a common denominator you can add the
tips
(cosx2+ sin2x)/(cosx2+
sin2x)
1/1 = 1

regards The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 sin(theta) = d, 
cos(theta) = a, 
tan(theta) = h, 
csc(theta) = f, 
sec(theta) = c, 
cot(theta) = g. solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))` tan−1{2cos(2sin−121​)}=tan−1{2cos(2×6π​)}=tan−1{2cos(3π​)}=tan−1{2×21​}=tan−1{1}=4π​ given that,l ‖ m and n cuts themlet,∠1 = 65o∠2 = x∠3 = 40o∠1 = ∠4 = 65o (alternate angle) (i)∠3 + ∠4 + ∠5 = 180o (angle sum property)40o + 65o + ∠5 = 180o∠5 = 75onow,∠2 + ∠5 = 180o (linear pair)x + 75o = 180ox = 105o [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] consider lhs,lhs = secθ (1 – sinθ) (secθ + tanθ)⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ - ) (secθ + tanθ)we know that = tanθ.⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ – tanθ) (secθ + tanθ)we know that (a + b) (a – b) = a2 – b2.⇒ secθ (1 – sinθ) (secθ + tanθ) = sec2θ – tan2θwe know that sec2θ – tan2θ = 1.∴ secθ (1 – sinθ) (secθ + tanθ) = 1 = rhshence proved. let a and b be the two given vectorsɵ = 150° is the angle between a and bif c is the resultant vector, then: ...1also, it is given that the resultant is perpendicular to one of the given vectors. let that given vector be anow, …2putting 2 in 1, we get:now, putting c = 10, ɵ = 150° ()so, hence, smaller vector is a = 20 units