Video summary

NBT MATH 2026 Preparation - Trigonometry (Part 1)

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Key takeaways

Educational

Main ideas / lessons conveyed (Trigonometry NBT exam prep)

The speaker works through 10 practice-style questions targeting core trigonometry skills likely to appear on the NBT 2021 exam, with emphasis on:

  • Reading angles and quadrants to determine the correct signs for sin/cos/tan
  • Converting between trig expressions using double-angle identities
  • Using standard formulas for the period of trig functions
  • Finding the range, including effects of vertical shifts
  • Simplifying expressions using negative-angle and cofunction identities (e.g., involving (90^\circ))
  • Applying graph transformations to interpret max/min values
  • Solving word problems using trig and/or the cosine rule

Detailed instruction-style content included

Question 1 (solve a cosine equation for (x) in correct quadrants)

Given: (y=\cos x), and the horizontal line (y=-0.5) intersects at points (A,B).

Steps:

  • Set ( \cos x = -0.5)
  • Use the reference angle:
    • (\cos^{-1}(0.5)=60^\circ)
  • Determine where cosine is negative:
    • Quadrant II: (x = 180^\circ-60^\circ = 120^\circ)
    • Quadrant III: (x = 180^\circ+60^\circ = 240^\circ)
  • Output coordinates:
    • (A=(120^\circ,-0.5))
    • (B=(240^\circ,-0.5))

Key lesson: Quadrant sign rules determine solutions beyond the reference angle.

Question 2 (simplify trig expression with negative angles + double-angle)

Method used:

  • Apply identities:
    • (\sin(-\theta)=-\sin\theta)
    • (\tan(-\theta)=-\tan\theta)
    • (\cos(-\theta)=\cos\theta)
  • Use double-angle form:
    • (\sin 2x = 2\sin x \cos x)
  • Carefully track negative signs through the algebra
  • Use a double-angle rewrite:
    • (1-2\sin^2 x) expressed as (-\cos 2x)

Result stated: [ -2\cos x-\cos 2x ] (then match to the correct multiple-choice option)

Question 3 (period of (3-\sin(4x)))

Given: (f(x)=3-\sin(4x))

Steps:

  • For (a\sin(bx)) or (a\cos(bx)), [ \text{Period}=\frac{360^\circ}{b} ]

  • Vertical shifts (like (+3) or (-1)) do not change the period

  • Here (b=4), so: [ \text{Period}=\frac{360^\circ}{4}=90^\circ ]

Question 4 (range of (f(x)=-1-2\sin(2x))-type function)

Given: (f(x)=-1-2\sin(2x)) (speaker describes shifting a sine graph down by 1)

Steps:

  • Base sine range:
    • If (y=2\sin(2x)), then (y\in[-2,2])
  • Apply vertical shift by (-1):
    • Minimum: (-2-1=-3)
    • Maximum: (2-1=1)

Range: [ -3 \le y \le 1 ]

Question 5 (period of (y=2\sin^2 x - 1))

Key transformation:

  • Use: [ \sin^2 x = \frac{1-\cos 2x}{2} ]

  • This leads to: [ 2\sin^2 x - 1 = -\cos 2x ]

Period:

  • Reduce to (\cos(2x)), so (b=2)
  • Therefore: [ \text{Period}=\frac{360^\circ}{2}=180^\circ ]

Question 6 (simplify a combined trig expression)

The expression includes terms like:

  • (\sin(x-180^\circ))
  • (-\sin(90^\circ+x))
  • (+\cos(-x))

Method:

  • Use negative-angle transformations and quadrant reasoning
  • Apply cofunction relationship near (90^\circ):
    • (\sin(90^\circ+\alpha)=\cos\alpha) (with sign handling)
  • Use (\cos(-x)=\cos x)

Simplification outcome stated: [ -\sin x ] (then choose from options)

Question 7 (graph shift: (\sin x) shifted (90^\circ) right)

Given: (f(x)=\sin x) shifted (90^\circ) to the right

Steps:

  • Shifting right by (90^\circ): (x \to x-90^\circ)

  • So: [ f(x)=\sin(x-90^\circ) ]

  • Use identity: [ \sin(x-90^\circ)=-\cos x ]

Final answer stated: [ -\cos x ]

Question 8 (find (\cos x) using (\sin 2x=y) and (\sin x=z))

Given:

  • (\sin 2x=y)
  • (\sin x=z)

Steps:

  • Use double-angle: [ \sin 2x = 2\sin x\cos x ]

  • Substitute: [ y = 2(z)\cos x ]

  • Solve: [ \cos x = \frac{y}{2z} ]

(then select the corresponding multiple-choice option)

Question 9 (max value of (k=\sin(x/2)+1) on ([-90^\circ,90^\circ]))

Given:

  • (x\in[-90^\circ,90^\circ])
  • (k=\sin(x/2)+1)

Steps:

  • Convert the interval:
    • (x/2\in[-45^\circ,45^\circ])

  • Over this interval: [ \sin(x/2)\in\left[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right] ]

  • Therefore:

    • Minimum: [ 1-\frac{1}{\sqrt2} ]

    • Maximum: [ 1+\frac{1}{\sqrt2} ]

Maximum stated: [ k_{\max}=1+\frac{1}{\sqrt2} ] (then rationalization is used as needed to match choices)

Question 10 (ladder slipping + distance between cyclists via cosine rule)

This question has two separate parts.

10(a) Ladder against a wall

Given:

  • Ladder against a 4 m wall
  • Initial angle of inclination is (30^\circ)
  • Ladder slips 1 m down, so wall height becomes (3) m

Steps:

  • Ladder length (L): [ \sin 30^\circ=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{4}{L} \Rightarrow L=\frac{4}{\sin 30^\circ}=\frac{4}{1/2}=8 ]

  • After slipping, new angle (\theta): [ \sin\theta=\frac{3}{8} \Rightarrow \theta=\sin^{-1}\left(\frac{3}{8}\right) ]

  • If (\theta) is not a special angle, leave it as (\sin^{-1}(3/8))

10(b) Distance between Sam and Tom after 15 minutes

Given:

  • Sam speed: 24 km/h (direction stated with an angle of (15^\circ) west of “not”—speaker’s wording implies a directional angle)
  • Tom speed: 32 km/h (direction described with (45^\circ))
  • Angle between their directions: (60^\circ)

Steps:

  • Time:
    • (15) minutes (= \frac{1}{4}) hour
  • Distances:
    • Sam: (24\times\frac{1}{4}=6) km
    • Tom: (32\times\frac{1}{4}=8) km

  • Use cosine rule for distance (d): [ d^2 = 6^2 + 8^2 - 2(6)(8)\cos 60^\circ ]

  • With (\cos 60^\circ=\frac12): [ d^2=36+64-96\cdot\frac12=100-48=52 ]

  • So: [ d=\sqrt{52}=\sqrt{4\cdot 13}=2\sqrt{13} ]

  • Choose from multiple-choice options

Speakers / sources featured

  • Primary speaker: The YouTube presenter/trainer (intro phrasing like “hi guys… welcome to my nbt training session”; no name given in the subtitles)
  • Source context mentioned: “NBT 2021 examination” and “past papers” (used to compile the 10 questions)

Original video